I’m trying to find the derivative of $x^2\sin x$ using only the limit definition of a derivative. I’ve tried two approaches, one using the difference quotient and another with the regular $x-a$ formula.
I’m stumped on both approaches and not sure where to go. Maybe I’m on the wrong track completely. The difference quotient gets messy quickly and I can’t figure out how to factor out $h$ to get it into a definable form. So then I tried: $$\frac{x^2\sin(x) – a^2\sin(a)}{x-a}.$$
Is it possible to apply the trig sum-to-product form to the numerator? I’m really just guessing playing around with identities trying to figure this out. Any tips would be appreciated!
Best Answer
\begin{align*} &\dfrac{(x+h)^{2}\sin(x+h)-x^{2}\sin x}{h}\\ &=\dfrac{(x+h)^{2}\sin(x+h)-x^{2}\sin(x+h)+x^{2}\sin(x+h)-x^{2}\sin x}{h}\\ &=\dfrac{((x+h)^{2}-x^{2})\sin(x+h)}{h}+\dfrac{x^{2}(\sin(x+h)-\sin x)}{h}\\ &=\dfrac{(2hx+h^{2})\sin(x+h)}{h}+\dfrac{x^{2}(\sin x\cos h+\cos x\sin h-\sin x)}{h}\\ &=(2x+h)\sin(x+h)+x^{2}\cdot\dfrac{\cos h-1}{h}+x^{2}\cos x\cdot\dfrac{\sin h}{h}\\ &\rightarrow 2x\sin x+x^{2}\cos x. \end{align*}