Let $\mathrm{f}(x)=x^2\sin x$. Using the definition \(\mathrm{g}'(x)=\lim_{h\to0}\dfrac{\mathrm{g}

Question

Let $\mathrm{f}(x)=x^2\sin x$. Using the definition $\mathrm{g}'(x)=\lim_{h\to0}\dfrac{\mathrm{g}(x+h)-\mathrm{g}(x)}{h}$, show from first principles that the derivative of $\mathrm{g}(x)=x^2$ is $2x$. Prove by induction that, for every positive integer $n$, $$\begin{aligned}\mathrm{f}^{(n)}(x)={}&[x^2-n(n-1)]\sin\left(x+\frac{n\pi}{2}\right)\&-2nx\cos\left(x+\frac{n\pi}{2}\right).\end{aligned}$$

Accepted Answer

(a) $\mathrm{g}'(x)=2x$ (b) $\mathrm{f}^{(n)}(x)=[x^2-n(n-1)]\sin(x+n\pi/2)-2nx\cos(x+n\pi/2)$, $n\ge1$

Quadratic sine derivative pattern

Quadratic sine derivative pattern

Solution:

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