### Differentiation(7.1-7.3)

#### Understand the concept of the derivative of a function

• Differentiate $$y = {x^4} + {x^2} + 3$$ with respect to x.

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

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

• (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 $$f{\kern 1pt} '(2)$$.
• (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

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