Question

Differentiate with respect to x
f(x) = sin^2x
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Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\:f(x) = {sin}^{2}x$

So,

$\rm :\longmapsto\:f(x + h) = {sin}^{2}(x + h)$

So,

By definition of First Principal, we have

$\rm :\longmapsto\:\boxed{ \tt{ \: f'(x) = \displaystyle\lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \: }}$

So, on substituting the values, we get

$\rm :\longmapsto\:f'(x) = \displaystyle\lim_{h \to 0} \frac{ {sin}^{2}(x + h) - {sin}^{2}x}{h}$

We know,

$\boxed{ \tt{ \: {sin}^{2}x - {sin}^{2}y = sin(x + y)sin(x - y) \: }}$

So, using this identity, we get

$\rm \:  =  \:\displaystyle\lim_{h \to 0} \frac{(sin(x + h) +sinx)(sin(x + h) - sinx)}{h}$

$\rm \:  =  \:\displaystyle\lim_{h \to 0}[sin(x + h) + sinx]\displaystyle\lim_{h \to 0} \frac{sin(x + h) - sinx}{h}$

We know,

$\boxed{ \tt{ \: sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg] \: }}$

So, using this, we get

$\rm \:  =  \:(sinx + sinx)\displaystyle\lim_{h \to 0} \frac{2cos\bigg[\dfrac{x + h + x}{2} \bigg]sin\bigg[\dfrac{x + h - x}{2} \bigg]}{h}$

$\rm \:  =  \:2 sinx\displaystyle\lim_{h \to 0} \frac{2cos\bigg[\dfrac{2x + h}{2} \bigg]sin\bigg[\dfrac{h}{2} \bigg]}{h}$

$\rm \:  =  \:4sinx\displaystyle\lim_{h \to 0}cos\bigg[\dfrac{2x + h}{2} \bigg]\displaystyle\lim_{h \to 0} \frac{sin\bigg[\dfrac{h}{2} \bigg]}{h}$

$\rm \:  =  \:4sinxcos\bigg[\dfrac{2x}{2} \bigg]\displaystyle\lim_{h \to 0} \frac{sin\bigg[\dfrac{h}{2} \bigg]}{\dfrac{h}{2} \times 2}$

$\rm \:  =  \:2sinxcosx\displaystyle\lim_{h \to 0} \frac{sin\bigg[\dfrac{h}{2} \bigg]}{ \dfrac{h}{2} }$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: \displaystyle\lim_{x \to 0} \frac{sinx}{x} = 1 \: }}$

So, using this, we get

$\rm \:  =  \:2sinxcosx \times 1$

$\rm \:  =  \:sin2x$

Hence,

$\rm \implies\:\boxed{ \tt{ \: \dfrac{d}{dx} {sin}^{2}x = sin2x \: }}$

More to know :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$