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 $\sin x$?
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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.
choco_addicted
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$y=e^{x\cdot sinx}$, 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: $y'=e^{xsinx}(x\cdot cosx+sinx)$
Itakura
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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 $$y'=\big(x \cos(x)+\sin(x)\big)e^{x\sin x}$$
Claude Leibovici
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I'm not sure if this method is a really good one since it's applicable in this specific situation only ! – Surb Mar 05 '16 at 10:44
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1@Surb. I am not sure I understand your point. As long as you can take logarithms, the method applies. It is, to me, the easiest way to explain the derivative of a product or a quotient. – Claude Leibovici Mar 05 '16 at 13:30
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Take $y=cosh(x)$, then $\ln(y)=\ln(\cosh(x))$ and thus $\frac{y'}{y}=\frac{d}{dx}(\ln(\cosh(x))$, and we arrive at the same problem than the beginning, i.e. to derivate a composition. – Surb Mar 05 '16 at 13:48
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it is the chain and the power rule: $${e^{x\sin(x)}}'=e^{x\sin(x)}(\sin(x)+x\cos(x))$$
Dr. Sonnhard Graubner
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$f(g(x))' = f'(g(x)) g'(x)$
– shilov Mar 05 '16 at 09:52