{"id":3165,"date":"2023-10-04T12:16:01","date_gmt":"2023-10-04T12:16:01","guid":{"rendered":"http:\/\/localhost\/ecole9ja\/?p=3165"},"modified":"2023-10-04T12:23:00","modified_gmt":"2023-10-04T12:23:00","slug":"week-2-ss2-second-term-further-mathematics-notes","status":"publish","type":"post","link":"https:\/\/ecolebooks.com\/nigeria\/posts\/week-2-ss2-second-term-further-mathematics-notes\/","title":{"rendered":"Week 2 – SS2 Second Term Further Mathematics Notes"},"content":{"rendered":"

\u00a0WEEK TWO
\n<\/strong>TOPIC: RULES OF DIFFERENTIATION
\n<\/strong>Derivative of Sum
\n<\/strong>Let f, U <\/em>and V<\/em> be functions of x such that
\nf<\/em>(x) = U<\/em>(x) + V<\/em>(x)
\nf<\/em>(x + \"\"\/x) = U<\/em>(x + \"\"\/x) + V<\/em>(x + \"\"\/x)
\nTherefore f <\/em>(x + \"\"\/x) \u2013 f<\/em> (x) = {U<\/em>(x + \"\"\/x) + V<\/em>(x + \"\"\/x) – U<\/em> (x) \u2013 V<\/em>(x)}
\n = U <\/em>(x) + \"\"\/x) \u2013 U<\/em>(x) + V<\/em>(x + \"\"\/x) \u2013 V<\/em>(x)
\n\"\"\/\"\"\/\"\"\/Therefore f <\/em>(x + \"\"\/x) \u2013 f<\/em> (x) = U<\/em>(x + \"\"\/x) \u2013 U<\/em>(x) + V<\/em>(x + \"\"\/x) \u2013 V<\/em>(x)
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"\/x \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"\/x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"\/x
\n\"\"\/\"\"\/\"\"\/Lim f <\/em>(x + \u2206x) \u2013 f <\/em>(x) = Lim U<\/em>(x + \u2206x) \u2013 U<\/em>(x) + LimV<\/em>(x + \u2206x) \u2013 V<\/em>(x)
\n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u2206x \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"\/x\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"\/x
\nTherefore f<\/em>‘ (x) = U<\/em>‘ (x) + V<\/em>‘(x)
\nIn other words, if y = U + V, where U and V are functions of x, then:
\n = +
\nHence, the derivation of a sum is the sum of the derivatives.
\nExamples
\nFind the derivative of each of the following<\/p>\n

    \n
  1. 2x3<\/sup> \u2013 5x2<\/sup> + 2\n<\/li>\n
  2. 3x2<\/sup> +\n<\/li>\n
  3. \n
    x3<\/sup> + 2x2<\/sup> + 1\n<\/div>\n

    \"\"\/x\n<\/li>\n

  4. \n
    + – 3\n<\/div>\n

    Solution<\/p>\n

      \n
    1. \n
      Let y = 2x3<\/sup> \u2013 5x2 <\/sup>+ 2\n<\/div>\n

      = 6x2<\/sup> \u2013 10x\n<\/li>\n

    2. \n
      Let y = 3x2<\/sup> +\n<\/div>\n

      = 3x2<\/sup> +
      \n= 6x –\n<\/li>\n

    3. \n
      let y = x3<\/sup> + 2x2<\/sup> + 1\n<\/div>\n

      \"\"\/ x= x2<\/sup> + 2x +
      \n = 2x + 2 –\n<\/li>\n

    4. \n
      y = + 1 – 3\n<\/div>\n

      = x + x -3\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

      = x – x
      \n= 1 – 1 <\/p>\n

      \u00a0Functions of a Function<\/strong>
      \n\t\tSuppose we know that y is a function of u and the u itself is also a functions of x, how do we find the derivation of y with respect to x?
      \nIn other words, given y = f (u) and u = h(x)
      \nWhat is ? By simple re \u2013 arrangement we can write
      \n = x ; \"\"\/ u \"\"\/ 0, \"\"\/ x \"\"\/ 0
      \nLim = Lim x }
      \n\t\t= lim x lim <\/p>\n

      \u00a0\"\"\/\"\"\/As \"\"\/x \u00a0\u00a0\u00a0\u00a0 0, \"\"\/u \u00a0\u00a0\u00a0\u00a00
      \nSo we can
      \n\"\"\/\"\"\/as lim <\/p>\n

      \u00a0Thus
      \n\t\t\t\t<\/strong>\"\"\/\"\"\/ = x <\/p>\n

      \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x
      \nThis is called the chain rule for differentiation. <\/p>\n

      \u00a0Examples
      \n<\/strong>Find the derivative of each of the following: <\/p>\n

        \n
      1. \n
        y = (3x2<\/sup> \u2013 2)3<\/sup>\n\t\t\t\t<\/div>\n<\/li>\n
      2. \n
        y =\n<\/div>\n<\/li>\n
      3. \n
        y = )\n<\/div>\n<\/li>\n
      4. \n
        \"\"\/y = )3<\/sup>\n\t\t\t\t<\/div>\n

        \u00a0<\/li>\n

      5. \n
        \"\"\/2<\/sup>)3 <\/sup>\n\t\t\t\t<\/div>\n<\/li>\n<\/ol>\n

        solution<\/p>\n

          \n
        1. \n
          Given \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0y\u00a0\u00a0\u00a0\u00a0= (3x2<\/sup> \u2013 2)3<\/sup>\n\t\t\t\t<\/div>\n

          Let \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0u\u00a0\u00a0\u00a0\u00a0= 3x2<\/sup>– 2
          \nthen\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0y\u00a0\u00a0\u00a0\u00a0= u3<\/sup><\/p>\n

          \u00a0<\/li>\n<\/ol>\n

          \u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a06x
          \n\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
          \n\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a03u2 <\/sup> x 6x
          \n\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a018xu2
          \n\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a018x (3x2 <\/sup>\u2013 2)2
          \n<\/sup><\/p>\n

            \n
          1. \n
            \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0y\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0\n<\/div>\n<\/li>\n<\/ol>\n

            Let \u00a0\u00a0\u00a0\u00a0u\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a01-2x3<\/sup>
            \n\t\tthen\u00a0\u00a0\u00a0\u00a0y\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0u<\/p>\n

            \u00a0\"\"\/ = u \u2013 \u00bd <\/sup><\/p>\n

            \u00a0\"\"\/=
            \n\t\t\t<\/sup>= – 6x2<\/sup>
            \n\t\t\"\"\/\"\"\/ = x
            \n\"\"\/\"\"\/\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
            \n\"\"\/ = ) \"\"\/<\/p>\n

              \n
            1. \n
              \"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0y\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a03<\/sup>\n\t\t\t\t<\/div>\n

              \"\"\/Let \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0u\u00a0\u00a0\u00a0\u00a0= \u00a0\u00a0\u00a0\u00a06 \u2013 x2
              \n<\/sup>\"\"\/then\u00a0\u00a0\u00a0\u00a0y\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0
              \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a05u-3<\/sup>
              \n\t\t\t\t\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0– 15u-4<\/sup>
              \n\t\t\t\t\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= – 2x
              \n\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0x\u00a0\u00a0\u00a0\u00a0
              \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= -15u-4 <\/sup> x -2x
              \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0=30 x u-4
              \n<\/sup>\u00a0\u00a0\u00a0\u00a0<\/sup>=\u00a0\u00a0\u00a0\u00a0(6-x2<\/sup>)4\"\"\/\"\"\/<\/sup>\n\t\t\t\t<\/li>\n

            2. \n
              \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0y\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0)\n<\/div>\n

              Let \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0u\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a01 + x2<\/sup><\/p>\n

              \u00a0then\u00a0\u00a0\u00a0\u00a0y\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 1
              \n\"\"\/\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= u \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
              \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = – u
              \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a01
              \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= –
              \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= 2x<\/p>\n

              \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = x
              \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 -1
              \n = x 2x
              \n\"\"\/ -x
              \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0= <\/p>\n

              \u00a0EVALUATION<\/strong>
              \n\t\t\t\tFind the derivative of the followings:
              \n(i) y = 8x5<\/sup> + 6x \u2013 7
              \n(ii) y = ( 4x3<\/sup> \u2013 3)4<\/sup>
              \n\t\t\t\t(iii) y = 3x2<\/sup> + 1\/x3<\/sup> + 2\/x
              \nThe Derivation of a Product
              \n<\/strong>We shall now consider the derivative of y = uv where u and v are functions of x.
              \n\u00a0\u00a0\u00a0\u00a0Let y\u00a0\u00a0\u00a0\u00a0 = \u00a0\u00a0\u00a0\u00a0uv
              \nThen y + \"\"\/y =\u00a0\u00a0\u00a0\u00a0(u + \"\"\/u) (v + \"\"\/v)
              \n = uv + u\"\"\/v + v\"\"\/u + \"\"\/u\"\"\/v \u2013 uv
              \n = u\"\"\/v u + \"\"\/u \"\"\/v
              \n = u + v+ \"\"\/u
              \n = u + v
              \nExamples
              \nFind the derivative of each of the following<\/p>\n

                \n
              1. \n
                y = (3 + 2x) (1 \u2013 x)\n<\/div>\n<\/li>\n
              2. \n
                y = (1 \u2013 2x + 3x2<\/sup>) (4 \u2013 5x2<\/sup>)\n<\/div>\n<\/li>\n
              3. \n
                y = (1 + 2x)2<\/sup>\n\t\t\t\t\t\t<\/div>\n<\/li>\n
              4. \n
                y = x3<\/sup> (3 \u2013 2x + 4x2<\/sup>)\n<\/div>\n

                Solution<\/p>\n

                  \n
                1. \n
                  y = (3 + 2x) (1 \u2013 x)\n<\/div>\n

                  Let u = 3 + 2x; v = 1 \u2013 x
                  \n = u + v
                  \n= (3 + 2x) x \u2013 1 + (1 \u2013 x) x 2
                  \n= -(3 + 2x) + 2(1 \u2013 x)
                  \n= -3 -2x + 2 \u2013 2x
                  \n= – 1 \u2013 4x\n<\/li>\n

                2. \n
                  \u00a0\u00a0\u00a0\u00a0y = (1 \u2013 2x + 3x2<\/sup>)(4 \u2013 5x2<\/sup>)\n<\/div>\n

                  Let u + 1 \u2013 2x + 3×2; v =4 \u2013 5 x2
                  \n<\/sup>\u00a0\u00a0\u00a0\u00a0= \u00a0\u00a0\u00a0\u00a0-2 + 6x; <\/sup>= -10x
                  \n\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0 u + v
                  \n\t\t\t\t\t\t\t\t\t\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0(1 -2x + 3x2)<\/sup> x (-10x)
                  \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0+ (4 \u2013 5x2<\/sup>)\n<\/li>\n

                3. \n
                  \u00a0\u00a0\u00a0\u00a0y\u00a0\u00a0\u00a0\u00a0=\u00a0\u00a0\u00a0\u00a0 (1 + 2x)2<\/sup>\n\t\t\t\t\t\t\t\t<\/div>\n

                  \"\"\/Let u \u00a0\u00a0\u00a0\u00a0= \u00a0\u00a0\u00a0\u00a0; v = (1 + 2x)2
                  \n<\/sup>
                  \n\u00a0\"\"\/= ; = 4(1 + 2x)
                  \n\u00a0\u00a0\u00a0\u00a0= u + v
                  \n = 4(1 + 2x) + (1 + 2x)2<\/sup> x 1
                  \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
                  \n 4 (1 + 2x) + (1 + 2x)2
                  \n<\/sup>\"\"\/\n\t\t\t\t\t\t\t\t<\/li>\n

                4. \n
                  y = x3<\/sup> (3 \u2013 2x + 4x2<\/sup>)
                  \n\t\t\t\t\t\t\t\t\t<\/sup><\/div>\n

                  Let u = x3<\/sup>; v = (3 \u2013 2x + 4x2<\/sup>)
                  \n = <\/sup>3x2<\/sup>; = ( – 2 + 8x) x (3 \u2013 2x + 4x2<\/sup>)
                  \n = <\/sup>3x2<\/sup>; =
                  \n = u + v<\/p>\n

                  \u00a0\"\"\/=\u00a0\u00a0\u00a0\u00a0 x3<\/sup> (4x \u2013 1)\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                  \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(3x \u2013 2x + 4x2) \u00bd + <\/sup>3x2<\/sup>
                  \n\t\tx (3 \u2013 2x + 4x2) \u00bd
                  \n<\/sup>\"\"\/= x3<\/sup> (4x -1) + 3x2 <\/sup> (3 -=2x + 4x2 )
                  \n<\/sup>(3x \u2013 2x + 4x2 <\/sup> ) \u00bd <\/p>\n

                  \u00a0The Derivative of a Quotient
                  \n<\/strong>Let y = , where u and v are functions of x and v \"\"\/ 0.
                  \ny + \"\"\/y = \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0
                  \n\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\"\"\/y = = v – u
                  \nExamples
                  \nFind the derivative of each of the following:<\/p>\n

                    \n
                  1. \n
                    \n\t\t\t\t<\/div>\n<\/li>\n
                  2. \n
                    3 + 2x \u2013 x2<\/sup>\n\t\t\t\t<\/div>\n

                    \"\"\/\n\t\t\t\t<\/li>\n

                  3. \n
                    \"\"\/ 2 + x\n<\/div>\n

                    x2<\/sup> + 2x + 7\n<\/li>\n

                  4. \n
                    \n\t\t\t\t<\/div>\n

                    \u00a0Solution<\/p>\n

                      \n
                    1. \n
                      \"\"\/y = 1 + x2<\/sup>\n\t\t\t\t\t\t<\/div>\n

                      1 \u2013 x2
                      \n<\/sup>Let u = 1 + x2<\/sup>; v = 1 \u2013 x2<\/sup>
                      \n\t\t\t\t\t\t = 2x; = – 2x
                      \n = v- u
                      \n= (1 \u2013 x2<\/sup>) x (2x) \u2013 (1 + x2<\/sup>) x (- 2x)
                      \n\"\"\/\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(1 \u2013 x2<\/sup>)2<\/sup>
                      \n\t\t\t\t\t\t= 2x \u2013 2x3<\/sup> + 2x + 2x3<\/sup>
                      \n\t\t\t\t\t\t\u00a0\u00a0\u00a0\u00a0 (1 \u2013 x2<\/sup>)
                      \n= 4x
                      \n\"\"\/ (1 \u2013 x2<\/sup>)2
                      \n<\/sup><\/li>\n

                    2. \n
                      y = 3 + 2x \u2013 x2<\/sup>\n\t\t\t\t\t\t<\/div>\n

                      \"\"\/
                      \n\t\t\t\t\t\tLet u = 3 + 2x \u2013 x2<\/sup>; = (1 \u2013 x)
                      \n = 2 \u2013 2x; = (1 + x)
                      \n = (1 + x) x 2 (1 \u2013 x) \u2013 (3 + 2x \u2013 x2<\/sup>) x
                      \n\"\"\/
                      \n\t\t\t\t\t\t= 4(1 + x)(1 \u2013 x) \u2013 (3 + 2x \u2013 x2<\/sup>)
                      \n\"\"\/ 2(1 + x)(1 + x)
                      \n\"\"\/= 4 \u2013 4x2<\/sup> \u2013 3 \u2013 2x + x2<\/sup>
                      \n\t\t\t\t\t\t 2(1 + x)
                      \n= 1 \u2013 2x \u2013 3x2<\/sup>
                      \n\t\t\t\t\t\t 2(1 + x)\n<\/li>\n

                    3. \n
                      \"\"\/y = 2 + x\n<\/div>\n

                      x2<\/sup> + 2x + 7
                      \nPut u = 2 + x; v = x2<\/sup> + 2x + 7
                      \n = 1; = 2x + 2
                      \n\"\"\/ = (x2<\/sup> + 2x + 7) x 1 \u2013 (2 + x)(2x + 2)
                      \n (x2<\/sup> + 2x + 7)2<\/sup>
                      \n\t\t\t\t\t\t= x2<\/sup> + 2x + 7 \u2013 2(x + 2)(x + 1)
                      \n\"\"\/ (x2<\/sup> + 2x + 7)2<\/sup>
                      \n\t\t\t\t\t\t= x2<\/sup> + 2x + 7 \u2013 2x2<\/sup> \u2013 6x \u2013 4
                      \n\"\"\/ (x2<\/sup> + 2x + 7)2
                      \n<\/sup>= x2<\/sup> + 2x + 7 \u2013 2x2<\/sup> \u2013 6x \u2013 4
                      \n\"\"\/ (x2<\/sup> + 2x + 7)2
                      \n<\/sup>\"\"\/ = – x2<\/sup> \u2013 4x + 3
                      \n (x2<\/sup> + 2x + 7)2<\/sup>\n\t\t\t\t\t\t<\/li>\n<\/ol>\n<\/li>\n<\/ol>\n

                      \u00a0HIGHER DERIVATIVES OF THE SECOND AND THIRD ORDER . DIFFERENTIATION OF IMPLICIT FUNCTIONS
                      \n\t\t\t\t<\/strong>HIGHER DERIVATIVES
                      \nGiven that y = f(x), is also a function of x.
                      \nThe derivative of with respect to x is
                      \n. is called the second derivative of y with respect to x, and it is usually denoted
                      \n\"\"\/
                      \n\t\t\u00a0\u00a0\u00a0\u00a0(read. Dee two y dee x squared).<\/p>\n

                      \u00a0\"\"\/Since \u00a0\u00a0\u00a0\u00a0 is also a function of x, successive derivatives can be found.<\/p>\n

                      \u00a0The third derivative of y with respect to x
                      \n\"\"\/\"\"\/\"\"\/Is and is written for short as d3<\/sup>y
                      \n dx3
                      \n<\/sup>S I <\/strong>d4<\/sup>y is the fourth derivative of y with respect to x.
                      \n dx4
                      \n<\/sup>in general is the nth derivative of y with respect to x.
                      \nWe recall that if y = f(x), is sometimes denoted f1<\/sup> (x).
                      \nSimilarly , , , are sometimes
                      \nDenoted fn<\/sup>(x), fm<\/sup>(x), fiv<\/sup> (x) respectively.
                      \nExample 22
                      \nFind the first second and third derivatives of each of the following:<\/p>\n

                        \n
                      1. 3×4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(b) 3x5\"\"\/<\/sup>2x4<\/sup> + x2\"\"\/<\/sup> 1\n<\/li>\n<\/ol>\n

                        (c) Inx\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0(d) ex4<\/sup>
                        \n\t\t(e) sin3x2<\/sup>
                        \n\t\tSolution:
                        \n<\/strong>(a)Let y = 3x4<\/sup>
                        \n\t\tThen = 12x3<\/sup>
                        \n\t\td2<\/sup>y = 36x2
                        \n<\/sup>dx2
                        \n<\/sup>d3<\/sup>y = 72x
                        \ndx3<\/sup><\/p>\n

                          \n
                        1. \n
                          Let y = 3x5<\/sup> \u2013 2x4<\/sup> + x2<\/sup> -1\n<\/div>\n

                          = 15x4<\/sup> \u2013 8x3<\/sup> + 2x
                          \nd2<\/sup>y = 60x3 <\/sup>– 24x2 <\/sup>+ 2
                          \ndx2
                          \n<\/sup><\/li>\n<\/ol>\n

                          d3<\/sup>y= 180x2<\/sup> \u2013 48x
                          \n dx3<\/sup><\/p>\n

                            \n
                          1. \n
                            Let y = Ink\n<\/div>\n

                            =
                            \n\t\t\t\td2<\/sup>y =
                            \ndx2
                            \n<\/sup>d3<\/sup>y = 2
                            \n\t\t\t\tdx3<\/sup>x3
                            \n<\/sup><\/li>\n

                          2. \n
                            Lat y = ex4<\/sup>\n\t\t\t\t<\/div>\n

                            = 4x3 <\/sup> ex4
                            \n<\/sup><\/li>\n<\/ol>\n

                            d2<\/sup>y = 12x2<\/sup>ex4 <\/sup> + 4x3<\/sup> ex4 <\/sup> 4x3
                            \n<\/sup> dx2
                            \n<\/sup>= 12x2<\/sup>ex4<\/sup>+ 16x6 <\/sup> – ex4
                            \n<\/sup>d3<\/sup>y= 24 ex4 <\/sup> + 12x2 <\/sup> (ex4<\/sup>) +96x5<\/sup> ex<\/sup>
                            \n\t\t dx3
                            \n<\/sup>ex4 <\/sup>+ 16x6 <\/sup> (ex4<\/sup>)
                            \n = 24 ex4<\/sup> + 48x5 <\/sup>ex4<\/sup> + 96x5\"\"\/<\/sup> ex4 <\/sup>+ 64x9 <\/sup> ex4
                            \n<\/sup><\/p>\n

                              \n
                            1. Let y = sin3x2<\/sup>\n\t\t\t<\/li>\n<\/ol>\n

                              = cos3x2 <\/sup> (3x2<\/sup>)
                              \n\t\t = 6xcos3x2<\/sup>
                              \n\t\td2<\/sup>y = 6cos 3x2<\/sup> \u2013 6x sin 3x2 <\/sup> (3x2<\/sup>)
                              \n\t\t dx2
                              \n<\/sup>= 6cos 3x2<\/sup> \u2013 36x2<\/sup> sin3x2
                              \n<\/sup>d3<\/sup>y = -6sin3x2 <\/sup> (3x2<\/sup>) \u2013 72x
                              \n dx3 <\/sup>
                              \n\t\t\"\"\/ sin3x2 <\/sup>36x2<\/sup>cos3x2 <\/sup> (3x2<\/sup>)
                              \n = -(6sin3x2<\/sup>) 6x \u2013 17xsin3x2<\/sup>– (36x2 <\/sup>cosx2<\/sup>) \"\"\/ 6x
                              \n= -36xsin3x2 <\/sup>-72xsin3x2<\/sup>– \u00a0\u00a0\u00a0\u00a0216x3<\/sup>cos3x2<\/sup>
                              \n\t\t = -108xsin3x2<\/sub>\u2013 216x3<\/sup> cos3x2<\/sup>
                              \n\t\t = -108x (sin3x2<\/sup> + <\/sub>2x2<\/sup>cos3x2<\/sup>)
                              \nEVALUATION
                              \n<\/strong>Find the second and third derivatives of (1) cos 6x ( 2) 4x5<\/sup> -5x
                              \nImplicit Differentiation
                              \n<\/strong>So far, we have treated relations. Of the form y = f (x). Examples of such relations are y = 3x2<\/sup> \u2013 2x + 1, y = 1 +
                              \nIn any of these relations, y is said to be expressed explicitly in terms of x. The derivative of y with respect to x can be found from the rules of differentiation which have been discussed in the previous units.
                              \nSometimes, the relationship between y and x may not be expressed explicitly.
                              \nFor example, consider x2<\/sup>y + xy3<\/sup> + 3x = 0. Here, the relation between y and x is not expressed explicitly. The relationship between y and x is said to be implicit.
                              \n<\/strong>In differentiating x2<\/sup>y+xy3<\/sup>+3x=0, y is treated as if it is a function of x and the rules of differentiation are applied in the appropriate manner. The process of differentiating implicit function is called implicit differentiation. <\/strong><\/p>\n

                              \u00a0Examples<\/strong>
                              \n\t\tDifferentiate each of the following implicitly:<\/p>\n

                                \n
                              1. X2<\/sup> + y2<\/sup> = 4\n<\/li>\n
                              2. X2<\/sup>y + y2 <\/sup>+ 4x = 1\n<\/li>\n
                              3. 4y2<\/sup>x \u2013 5x2<\/sup>y3+ 4y = 0\n<\/li>\n
                              4. \n
                                (x + y)2<\/sup> = 5\n<\/div>\n

                                Solution
                                \n<\/strong><\/li>\n

                              5. \n
                                X2<\/sup> + y2<\/sup> = 4
                                \n\t\t\t\t\t<\/strong><\/div>\n

                                Differentiating term by term with respect to x:
                                \n2x + 2y = 0
                                \n 2y = -2x
                                \n =
                                \n=\n<\/li>\n

                              6. \n
                                X2<\/sup> + y2<\/sup>x + 4x = 1
                                \n\t\t\t\t\t<\/strong><\/div>\n

                                2xy + x2<\/sup> + 2yx + y2<\/sup> + 4 = 0
                                \n(x2<\/sup> + 2yx) = -y2<\/sup> 2xy \u2013 4
                                \n = -(y2<\/sup> + 2xy + 4)
                                \n\t\t\t\tX2 <\/sup>+ 2yx\n\t\t\t\t\t<\/li>\n

                              7. \n
                                4y2<\/sup>x -5x2<\/sup>y2<\/sup> + 4y = 0\n<\/div>\n

                                8yx + 4y2 <\/sup>– 15x2<\/sup>y2<\/sup> – 10xy3<\/sup> + 4 = 0
                                \n(8xy \u2013 15x2<\/sup>y2<\/sup> + 4) = 10xy3<\/sup> \u2013 4y2
                                \n<\/sup>= 10xy3<\/sup>– 4y2<\/sup>
                                \n\t\t\t\t\t8xy \u2013 15x2<\/sup>y2<\/sup> + 4\n<\/li>\n

                              8. \n
                                (x + y)2<\/sup> = 5\n<\/div>\n

                                \"\"\/x2<\/sup> + 2xy + y2 <\/sup> = 5
                                \n2x + 2x +2y+2y = 0
                                \n (2x + 2y) = -2x-2y
                                \n =
                                \n =
                                \n = = -1\n<\/li>\n<\/ol>\n

                                \u00a0EVALUATION
                                \n<\/strong>Differentiate the followings ;
                                \n(i) y= (3x+4) (6x-8)
                                \n(ii) y = 6x+7\/2x-3<\/p>\n

                                \u00a0GENERAL EVALUATION<\/strong>
                                \n\t\t1) Differentiate y = ( 7x4<\/sup> \u2013 6 )5<\/sup>
                                \n\t\t2) Differentiate y = ( 2x + 5) ( 6x \u2013 8)
                                \n3) Find the derivative of y = 3x2<\/sup> \u2013 5\/x + 3
                                \n4) Find the derivative of y = 8\/ ( 9 \u2013 x5<\/sup>)4<\/sup>
                                \n\t\t5) Find the derivative of y = 2x4<\/sup> -5x3<\/sup> -+ 6
                                \n6) If x3<\/sup>– y2<\/sup> + 6xy = 0 find dy\/dx
                                \n7) Find d3<\/sup>y\/dx3<\/sup> given that y = 8x5<\/sup> \u2013 3x4<\/sup> + 9x3<\/sup> -7x2<\/sup> +6x+4<\/p>\n

                                \u00a0Reading Assignment
                                \n<\/strong>New Further MathsProject 2 page 121 \u2013 126<\/p>\n

                                \u00a0WEEKEND ASSIGNMENT<\/strong>
                                \n\t\t1) If y = 3x4<\/sup> -7x + 5 find dy\/dx a) 12x3<\/sup> b) 12x3<\/sup> \u2013 7 c) 12x3<\/sup> + 5 d) 12x3<\/sup> + 12
                                \n2) Find the second derivative of cos 5x a) 5sin5x b) -25cos5x c) 25cos5x d) -25sin5x
                                \n\t\t\t3) 2) If x2<\/sup>y + 4xy =1 find dy\/dx a) 4+2xy\/x2<\/sup> b) 4-2xy\/x2<\/sup> c) -4-2xy\/x2<\/sup> d) -4+2xy\/x
                                \n4) Given that y = x2<\/sup> + 3x + 2, find dy\/dx at x = 2 a) 6 b) 4 c) 7 d) 5
                                \n5) Given that y = ( 2x + 3)4<\/sup> find dy\/dx a) 18(2x + 3)3<\/sup> b) 4(2x + 3)4<\/sup> c) 8(2x + 3)3<\/sup> d) 2( 2x+3)3<\/sup><\/p>\n

                                \u00a0THEORY
                                \n<\/strong>1) Differentiate y = (2x2<\/sup> -3)3<\/sup>\/x
                                \n2) Differentiate y = (2x+ 3)3<\/sup> (4x2<\/sup> -1)2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

                                \u00a0WEEK TWO TOPIC: RULES OF DIFFERENTIATION Derivative of Sum Let f, U and V be…<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1,253],"tags":[],"class_list":["post-3165","post","type-post","status-publish","format-standard","hentry","category-posts","category-second-term-ss2-further-mathematics"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3165","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/comments?post=3165"}],"version-history":[{"count":1,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3165\/revisions"}],"predecessor-version":[{"id":3166,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/posts\/3165\/revisions\/3166"}],"wp:attachment":[{"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/media?parent=3165"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/categories?post=3165"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ecolebooks.com\/nigeria\/wp-json\/wp\/v2\/tags?post=3165"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}