Find the of derivative of the function from the first principle: …

Question

Find the of derivative of the function from the first principle: $\sin x^2$.

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Answer

Let $y=\sin x^2$
Then, $y+\delta y=\sin (x+\delta x)^2$
$\Rightarrow \delta y=\sin (x+\delta x)^2-\sin x^2$
Using first principle,
$\Rightarrow \frac{\delta y}{\delta x}=\frac{\sin (x+\delta x)^2-\sin x^2}{\delta x}$
$\Rightarrow \frac{d y}{d x}=\lim _{\delta x \rightarrow 0} \frac{\delta y}{\delta x}$
$=\lim _{\delta x \rightarrow 0} \frac{\sin (x+\delta x)^2-\delta \sin x^2}{\delta x}$
$=\lim _{\delta x \rightarrow 0} \frac{2 \cos \left[\frac{(x+\delta x)^2+x^2}{2}\right] \sin \left[\frac{(x+\delta x)^2-x^2}{2}\right]}{\delta x}$
${\left[\text { using }(\sin C-\sin D)=2 \cos \left(\frac{C+D}{2}\right) \sin \left(\frac{C-D}{2}\right)\right]}$
$=\lim _{\delta x \rightarrow 0} 2 \cos \left[\frac{(x+\delta x)^2+x^2}{2}\right] \frac{\sin \left[\left(x+\frac{\delta x}{2}\right) \cdot \delta x\right]}{\left(x+\frac{\delta x}{2}\right) \cdot \delta x}\left(x+\frac{\delta x}{2}\right)$
$=2 \cdot \lim _{\delta x \rightarrow 0} \cos \left[\frac{(x+\delta x)^2+x^2}{2}\right] \cdot \lim _{\delta x \rightarrow 0} \frac{\sin \left[\left(x+\frac{\delta x}{2}\right) \cdot \delta x\right]}{\left(x+\frac{\delta x}{2}\right) \cdot \delta x}$
$\cdot \lim _{\delta x \rightarrow 0}\left(x+\frac{\delta x}{2}\right)$
$=\left[2 \times \cos x^2 \times 1 \times x\right]=2 x \cos x^2$
Hence, $\frac{d}{d x}\left(\sin x^2\right)=2 x \cos x^2$

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