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Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$ $$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2}{\sin(x+h)}-\frac{x^2}{\sin x}}{h}$$ $$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2\sin x-x^2\sin(x+h)}{\sin(x+h)\sin x}}{h}$$ $$f'(x)=\lim_{h\to0}\frac{(x+h)^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)}$$ $$f'(x)=\lim_{h\to0}\frac{x^2}{\sin x}\frac{(1+\frac{h}{x})^2\sin x-\sin(x+h)}{h\sin(x+h)}$$ I am stuck here.

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    Why would anyone want to do this? – 2017-01-14
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    You can find a proof for quotient rule and plug in $f = g/h,g(x) = x^2,h(x) = \sin x$ to *every step*. – 2017-01-14
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    @ZacharySelk : Strange, I left a comment that this is time waste exercise but it disapeared, voting to close. – 2017-01-14
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    I'm voting to close this question as off-topic because it is an exercise in time wasting. – 2017-01-14

3 Answers 3

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\begin{eqnarray} \dfrac{f(x+h)-f(x)}{h}&=&\dfrac{1}{h}\left[\dfrac{(x+h)^2}{\sin(x+h)}-\dfrac{x^2}{\sin x}\right]\\ &=&\dfrac{(x+h)^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)}\\ &=&\dfrac{[(x+h)^2-x^2]\sin x+x^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)}\\ &=&\dfrac{[(x+h)^2-x^2]\sin x-x^2[\sin(x+h)-\sin x]}{h\sin x\sin(x+h)}\\ &=&\dfrac{1}{\sin x\sin(x+h)}\left[\dfrac{(x+h)^2-x^2}{h}\cdot\sin x-x^2\cdot\dfrac{\sin(x+h)-\sin x}{h}\right] \end{eqnarray} Since \begin{eqnarray} \lim_{h\to0}\dfrac{1}{\sin x\sin(x+h)}&=&\dfrac{1}{\sin^2x}\\ \lim_{h\to0}\dfrac{(x+h)^2-x^2}{h}&=&\lim_{h\to0}\dfrac{x^2+2xh+h^2-x^2}{h}=\lim_{h\to0}\dfrac{2xh+h^2}{h}=\lim_{h\to0}(2x+h)=2x\\ \lim_{h\to0}\dfrac{\sin(x+h)-\sin x}{h}&=&\dfrac{d}{dx}(\sin x)=\cos x \end{eqnarray} it follows that $$ \lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}=\dfrac{1}{\sin^2x}[2x\cdot\sin x-x^2\cdot\cos x]=\dfrac{2x\cdot\sin x-x^2\cdot\cos x}{\sin^2x} $$

Added: Using the limits $$ \lim_{h\to0}\dfrac{\cos h-1}{h^2}=-\frac12,\quad \lim_{h\to0}\dfrac{\sin h}{h}=1, $$ we get: \begin{eqnarray} \lim_{h\to0}\dfrac{\sin(x+h)-\sin x}{h} &=&\lim_{h\to0}\dfrac{\sin x\cos h+\cos x\sin h-\sin x}{h}\\ &=&\lim_{h\to0}\left[\dfrac{\sin x\cos h-\sin x}{h}+\dfrac{\cos x\sin h}{h}\right]\\ &=&\sin x\cdot\lim_{h\to0}\dfrac{\cos h-1}{h}+\cos x\cdot\lim_{h\to0}\dfrac{\sin h}{h}\\ &=&\sin x\cdot 0+\cos x\cdot 1\\ &=&\cos x \end{eqnarray}

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    If you can't beat them, fool them. `;-)` – 2017-01-14
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Note that $$\lim _{ h\to 0 } \frac { 2\sin ^{ 2 }{ \frac { h }{ 2 } } }{ h } =\lim _{ h\to 0 } \frac { \sin { \frac { h }{ 2 } \cdot \sin { \frac { h }{ 2 } } } }{ \frac { h }{ 2 } } =\lim _{ h\to 0 } \frac { \sin { \frac { h }{ 2 } } }{ \frac { h }{ 2 } } \sin { \frac { h }{ 2 } } =0$$ so $$ \begin{align} f'(x)&=\lim _{ h\to 0 } \frac { (x+h)^{ 2 }\sin x-x^{ 2 }\sin (x+h) }{ h\sin x\sin (x+h) }\\&=\lim _{ h\to 0 } \frac { { x }^{ 2 }\sin x+2xh\sin { x+{ h }^{ 2 }\sin { x } } -x^{ 2 }\sin x\cos { h } -{ x }^{ 2 }\sin { h\cos { x } } }{ h\sin x\sin (x+h) }\\ &=\lim _{ h\to 0 } \frac { { x }^{ 2 }\sin x\cdot \left( 1-\cos { h } \right) +h\sin { x\cdot \left( 2x+h \right) } -{ x }^{ 2 }\sin { h\cos { x } } }{ h\sin x\sin (x+h) }\\&=\lim _{ h\to 0 } \frac { { x }^{ 2 }\sin x\cdot \frac { 2\sin ^{ 2 }{ \frac { h }{ 2 } } }{ h } +\sin { x\cdot \left( 2x+h \right) } -{ x }^{ 2 }\frac { \sin { h } }{ h } \cdot \cos { x } }{ \sin x\sin (x+h) }\\& =\frac { 2x\sin { x-{ x }^{ 2 }\cos { x } } }{ \sin ^{ 2 }{ x } } \end{align}$$

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Let's go back to here:

$$f'(x)=\lim_{h\to0}\frac{(x+h)^2\sin x-x^2\sin(x+h)}{h\sin x\sin(x+h)} =$$ $$\lim_{h\to0}\frac{x^2(\sin x - \sin(x+h)) + (2xh+h^2)\sin x}{h\sin x\sin(x+h)} =$$ $$\lim_{h\to0}\left(\frac{x^2(\sin x - \sin x\cos h - \cos x\sin h)}{h\sin x\sin(x+h)}+\frac{2x+h}{\sin(x+h)}\right) =$$ $$\lim_{h\to0}\left(\frac{hx^2}{\sin(x+h)}\cdot\frac{1-\cos h}{h^2} -\frac{x^2\cos x}{\sin x\sin(x+h)}\cdot\frac{\sin h}{h}+\frac{2x+h}{\sin(x+h)}\right) =$$ $$0 - \frac{x^2\cos x}{\sin^2x} + \frac{2x}{\sin x} =$$ $$\frac{2x\sin x - x^2\cos x}{\sin^2x}$$