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Could you explain me how to modify: $$\frac{d}{dx}\left(e^{-7\sin(4x)\ln(x)}\right)$$ to this form: $$7x^{7sin(4x)} \cdot \left(4\cos(4x)\ln(x)+\frac{\sin(4x)}{x}\right).$$ Thank you :).

I apologize for the registration example. Next time I will try to write it correctly. Thank you for understanding

MrMazgari
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Marián Slovák
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  • The chain rule in this case says that the required derivative is equals to $$e^{-7\sin 4x \ln x}\frac{d}{dx}\left(-7\sin 4x \ln x\right)$$ Try working this out. – Yiyuan Lee Nov 09 '15 at 14:35

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First note that: $$e^{-7\sin(4x)\ln(x)}=\left(e^{\ln(x)}\right)^{-7\sin(4x)}=x^{-7\sin(4x)}.$$

By the chain rule and product rule, we have: \begin{align} \frac{d}{dx}\left(e^{-7\sin(4x)\ln(x)}\right)&=\left(-7\cos(4x)\ln(x)-\frac{7\sin(4x)}{x}\right)e^{-7\sin(4x)\ln(x)} \\ &=-7x^{-7\sin(4x)}\cdot\left(\cos(4x)\ln(x)+\frac{\sin(4x)}{x}\right). \end{align}

(I think you are missing a minus sign in your question.)

MrMazgari
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