Differentiation(7.1-7.3)

Understand the concept of the derivative of a function

  • Theory
  • Examples
    Differentiate \(y = {x^4} + {x^2} + 3\) with respect to x.
  • Solutions

    \(\begin{array}{1}\frac{{dy}}{{dx}} = \frac{d}{{dx}}({x^4} + {x^2} + 3)\\ = \frac{d}{{dx}}({x^4}) + \frac{d}{{dx}}({x^2}) + \frac{d}{{dx}}{(3)^{^{^{^{}}}}}\\ = 4{x^{4 - 1}} + 2{x^{2 - 1}} + {0^{^{^{}}}}\\ = {\underline{\underline {4{x^3} + 2x}} ^{}}\end{array}\)

Understand the addition rule, product rule, quotient rule and chain rule of differentiation

  • Theory
  • Examples
    (a) Let \(u = \sqrt {2 + x} \), find \(\frac{{du}}{{dx}}\).
    (b) Let \(y = \sqrt {2 + \sqrt {2 + x} } \), find \(\frac{{dy}}{{dx}}\).
    (c) Hence if \(f(x) = \sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt {2 + x} } } } } \), find \(f{\kern 1pt} '(2)\).
  • Solutions
    (a)
    \(\begin{array}{1}\frac{{dy}}{{dx}} = \frac{1}{2}{(2 + x)^{ - \,\frac{1}{2}}}\\
    = \underline{\underline {\frac{1}{{2\sqrt {2 + x} }}}} \end{array}\)



    (b)
    \(\begin{array}{1}y = \sqrt {2 + \sqrt {2 + x} } \\ = {\sqrt {2 + u} ^{}} \\\frac{{dy}}{{dx}} =
    \frac{{dy}}{{du}} \cdot {\frac{{du}}{{dx}}^{^{}}}\\ = \frac{1}{{2\sqrt {2 + u} }} \cdot {\frac{1}{{2\sqrt {2 + x} }}^{}}\\ = {\underline{\underline {\frac{1}{{4\sqrt {2 + x} \sqrt {2 + \sqrt {2 + x} } }}}} ^{}}\end{array}\)



    (c) Let \(p = \sqrt {2 + y} = \sqrt {2 + \sqrt {2 + \sqrt {2 + x} } } \), \(q = \sqrt {2 + p} = \sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt {2 + x} } } } \),
    \(r = \sqrt {2 + q} = \sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt {2 + \sqrt {2 + x} } } } } = f(x)\).

    \(\begin{array}{1}f{\kern 1pt}'(x) = \frac{{dr}}{{dx}}\\
    = \frac{{dr}}{{dq}} \cdot \frac{{dq}}{{dp}} \cdot \frac{{dp}}{{dy}} \cdot \frac{{dy}}{{du}} \cdot {\frac{{du}}{{dx}}^{^{}}}\\
    = \frac{1}{{2\sqrt {2 + q} }} \cdot \frac{1}{{2\sqrt {2 + p} }} \cdot \frac{1}{{2\sqrt {2 + y} }} \cdot \frac{1}{{2\sqrt {2 + u} }} \cdot {\frac{1}{{2\sqrt {2 + x} }}^{}}\end{array}\)
    When x = 2, \(u = y = p = q = 2\).
    \(\begin{array}{1}\therefore {\rm{}}f{\kern 1pt} '(2) = & \;\frac{1}{{32(2)(2)(2)(2)(2)}}\\
    =& \;\underline{\underline {\frac{1}{{1{\rm{ }}024}}}} \end{array}\)

Find the derivatives of functions involving algebraic functions, trigonometric functions, exponential functions and logarithmic functions

  • Theory
  • Examples
    If \(y = {(3\sin x + \cos 2x)^3}\), find \(\frac{{dy}}{{dx}}\).
  • Solutions
    \(\begin{array}{1}\frac{{dy}}{{dx}}
    = \frac{d}{{dx}}{(3\sin x + \cos 2x)^3}\\
    = \frac{d}{{d(3\sin x + \cos 2x)}}{(3\sin x + \cos 2x)^3} \cdot \frac{d}{{dx}}{(3\sin x + \cos 2x)^{^{^{^{}}}}}\\
    = 3{(3\sin x + \cos 2x)^2}{[3\cos x + ( - \sin 2x)(2)]^{^{^{}}}}\\
    = {\underline{\underline {3{{(3\sin x + \cos 2x)}^2}(3\cos x - 2\sin 2x)}} ^{^{}}}\end{array}\)