help needed to compute derivative of $e^{x\sin x}$

Question

How should I compute the derivative of $e^{x\sin x}$ ? I am a student of class 11, so can you explain me how to do this without high level mathematics ( I know first principles ) I know that derivative of $e^x$ is $e^x$, but I cannot understand what to do with that $\sin x$?

Answer 1

Remind the chain rule $$(f(g(x)))'=g'(x)f'(g(x))$$ and product rule: $$(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)$$ You can use follwing formulas: $$ (e^x)'=e^x\text{ (you already know)}\\ (\sin x)'=\cos x\\ (x\sin x)'=\sin x + x\cos x $$ and so $$ (e^{x\sin x})'=(\sin x+x\cos x)e^{x\sin x}. $$ Intuitive understanding of the derivatives of $\sin x$ and $\cos x$ may help you.

Answer 2

$y=e^{x\cdot sinx}$, let $u=x\cdot sinx$

$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$ (chain rule)

$\frac{dy}{du}=e^u$

$\frac{du}{dx}=x\cdot cosx + 1\cdot sinx$ (product rule)

$y'=e^u\cdot (x\cdot cosx+sinx)$

therefore: $y'=e^{xsinx}(x\cdot cosx+sinx)$

Answer 3

Hint

Logarithmic differentiation is very useful $$y=e^{x\sin( x)}$$ Take logarithms $$\log(y)=x \sin(x)$$ Differentiate $$\frac{y'}y=x \cos(x)+\sin(x)$$ So $$y'=\big(x \cos(x)+\sin(x)\big)e^{x\sin x}$$

Answer 4

it is the chain and the power rule: $${e^{x\sin(x)}}'=e^{x\sin(x)}(\sin(x)+x\cos(x))$$