Chapter 2 Rules of Differentiation(1/2)

Find $f'(x)$

1. $f(x) = \frac{1}{x^2}+\sqrt[3]{x^2}$.
2. $f(x) = \frac{e^x+x^2}{1-3e^x}$.
3. $f(x) = x^3sinx$.
4. $f(x) = \frac{x^2+x-2}{x^3+6}$.
5. $f(x) = \sqrt{x}g(x)$,where $g(4) = 2,g'(4) = 3$. find $f'(4)$

$f(x) = x^{-2}+x^{\frac{2}{3}}$,$f(x) = -2x^{-3}+\frac{2}{3}x^{-\frac{1}{3}}$

$f(x) = \frac{(e^x+2x)(1-3e^x)-(e^x+x^2)(-3e^x)}{(1-3e^x)^2} = \frac{(3x^2-6x+1)e^x+2x}{(1-3e^x)^2}$

$f(x) = x^3sinx$, $f'(x) = 3x^2sinx+x^3cosx$

$f(x) = \frac{x^2+x-2}{x^3+6}$, $f'(x) = \frac{(2x+1)(x^3+6)-(x^2+x-2)(3x^2)}{(x^3+6)^2}= \frac{-x^4-2x^3+6x^2+12x+6}{(x^3+6)^2}$

$\because f(x) = \sqrt{x}g(x) \therefore f'(x) = (\frac{d}{dx}x^{\frac{1}{2}})g(x) + x^{\frac{1}{2}}(\frac{d}{dx}g(x)) = \frac{-1}{2\sqrt{x}}\cdot g(x) + \sqrt{x} \cdot g'(x)$
$\Rightarrow f'(4) = \frac{1}{4}\cdot 2 + 2 \cdot 3 = \frac{13}{2}$

$h(x) = (x^2+1)^{100}$ Find $h'(x)$

Let $f(x) = x^{100}, \; g(x) = x^2+1$
$\Rightarrow h(x) = (f\circ g)(x) = f(g(x))$
$\frac{d}{dx}h(x) = \frac{d}{dx}(f(g(x))) = \frac{d}{d(x^2+1)}(x^2+1)^{100} \cdot \frac{d}{dx}(x^2+1) = 100(x^2+1)^{99}\cdot (2x) = 200x(x^2+1)^{99}$

$h(x) = sinx^2$ Find $h'(x)$
$k(x) = sin^2x$ Find $k'(x)$

Let $f(x) = x^2, \; g(x) = sinx$
$\Rightarrow k(x) = f(g(x))$
$\frac{d}{dx}k(x) = \frac{d}{dx}f(g(x)) = \frac{df(g(x))}{dg(x)}\cdot \frac{dg(x)}{dx} = \frac{d}{dsinx}(sinx)^2 \cdot \frac{d}{dx}(sinx) = 2(sinx)(cosx)$

Let $f(x) = sinx, \; g(x) = x^2$
$\Rightarrow h(x) = f(g(x))$
$\frac{d}{dx}h(x) = \frac{d}{dx}f(g(x)) = \frac{df(g(x))}{dg(x)}\cdot \frac{dg(x)}{dx} = \frac{d}{dx^2}sin(x^2) \cdot \frac{d}{dx}(x^2) = cos(x^2)\cdot 2x = 2xcosx^2$

Find $f'(x)$

1. $f(x) = \left ( \frac{3x-1}{x^2+3} \right )^2$
2. $f(x) = tan(x^2-1)^2$
3. $f(x) = sin(x^2+2x)$
4. $f(x) = \left\{\begin{matrix} x^2sin\frac{1}{x} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$
5. $f(x) = e^{sec^2x}$

Let $h(x) = x^2 k(x) = \frac{3x-1}{x^2+3} \Rightarrow f(x) = h(k(x))$
$\frac{d}{dx}f(x) = \frac{d}{dx}h(k(x)) = \frac{d}{d(\frac{3x-1}{x^2+3})}\left ( \frac{3x-1}{x+3} \right )^2 \cdot \frac{d}{dx}\left ( \frac{3x-1}{x^2+3} \right ) = 2(\frac{3x-1}{x^2+3}) \cdot \frac{3(x^2+3)-(3x-1)\cdot 2x}{(x^2+3)^2} = 2(\frac{3x-1}{x^2+3}) \cdot \frac{-3x^2+2x+9}{(x^2+3)^2}$

Let $h(x) = tanx, k(x) = x^2, r(x) = x^2-1 \Rightarrow f(x) = (h\circ k \circ r)(x) = h(k(r(x)))$
$\frac{d}{dx}f(x) = \frac{d}{dx}h(k(r(x))) = \frac{dh(k(r(x)))}{h(r(x))}\cdot \frac{dk(r(x))}{dr(x)}\cdot \frac{dr(x)}{dx} = \frac{d}{d(x^2-1)^2}tan(x^2-1)^2 \cdot \frac{d}{d(x^2-1)}(x^2-1)^2\cdot\frac{d}{dx}(x^2-1) = sec^2(x^2-1)^2\cdot 2(x^2-1) \cdot 2x = 4x(x^2-1)sec^2(x^2-1)^2$

$f'(x) = cos(x^2+2x)\cdot \frac{d}{dx}(x^2+2x) = (2x+2)cos(x^2+2x)$

If $x \neq 0$, $\frac{d}{dx}f(x)= \frac{d}{dx}(x^2sin\frac{1}{x}) = 2xsin\frac{1}{x}+x^2\frac{d}{dx}sin\frac{1}{x} = 2xsin\frac{1}{x}+x^2cos\frac{1}{x}\cdot \frac{d}{dx}(x^{-1}) = 2xsin\frac{1}{x} - cos\frac{1}{x}$.
If $x =0$, then $\lim_{h \to 0}\frac{f(0+h)-f(0)}{h} = \lim_{h \to 0}\frac{sin\frac{1}{h}}{\frac{1}{h}} = 0$

$\therefore f'(x) = \left\{\begin{matrix} 2xsin\frac{1}{x} - cos\frac{1}{x} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

Let $f_1(x) = e^x,f_2(x) = x^2,f_3(x) = secx$
$\Rightarrow f(x) = f_1(f_2(f_3(x)))$
$f'(x) = e^{sec^2x}\cdot 2(secx) \cdot \frac{d}{dx}(secx)= 2tanx \cdot sec^2x \cdot e^{sec^2x}$

$x^2+y^2 = 1$ $(a) (\frac{1}{2},\frac{\sqrt{3}}{2})$ $(b) (-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})$ 求 tangent line.

• a: $\frac{dy}{dx}|_{x=\frac{1}{2}} = \frac{-x}{\sqrt{1-x^2}}_{\frac{1}{2}} = \frac{-1}{\sqrt{3}} \therefore y-\frac{\sqrt{3}}{2} = -\frac{1}{\sqrt{3}}(x-\frac{1}{2})$
• b: $\frac{dy}{dx}|_{x=-\frac{1}{\sqrt{2}}} = \frac{-x}{\sqrt{1-x^2}}_{x=-\frac{1}{\sqrt{2}}} = -1 \therefore y+\frac{1}{\sqrt{2}} = -(x+\frac{1}{\sqrt{2}})$

• a: $\frac{dy}{dx}|_{(\frac{1}{2},\frac{\sqrt{3}}{2})} = \frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{-1}{\sqrt{3}}$
• b: $\frac{dy}{dx}|_{(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}})} = -\frac{-\frac{1}{\sqrt{2}}}{-\frac{1}{\sqrt{2}}} = -1$

$x^3+y^3 = 6xy$ Find $\frac{dy}{dx}$

$\frac{d}{dx}(x^3+y^3) = \frac{d}{dx}(6xy)$
$\Rightarrow 3x^2+3y^2\frac{dy}{dx} = 6(y+x\cdot \frac{dy}{dx})$
$\Rightarrow (3y^2-6x)\frac{dy}{dx} = 6y-3x^2$
$\Rightarrow \frac{dy}{dx} = \frac{6y-3x^2}{3y^2-6x}$if $3y^2-6x \neq 0$

$sin(x+y) = y^2cosx$ Find $y'$

$\frac{d}{dx}(sin) = \frac{d}{dx}(6xy)$
$\Rightarrow 3x^2+3y^2\frac{dy}{dx} = 6(y+x\cdot \frac{dy}{dx})$
$\Rightarrow (3y^2-6x)\frac{dy}{dx} = 6y-3x^2$
$\Rightarrow \frac{dy}{dx} = \frac{6y-3x^2}{3y^2-6x}$if $3y^2-6x \neq 0$

$sin(x+y) = y^2cosx$.Find $y'$

$\frac{d}{dx}sin(x+y) = \frac{d}{dx}y^2cosx$

$\Rightarrow cos(x+y) = \frac{d}{dx}(x+y) = 2yy'cosx + y^2(-sinx)$
$\Rightarrow cos(x+y) + y'cos(x+y) = 2yy'cosx-y^2sinx$
$\Rightarrow y' = \frac{y^2sinx+cos(x+y)}{2ycosx-cos(x+y)}$ if $2ycosx-cos(x+y)\neq 0$