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How do I differentiate $5^{x \cos x}$? From my book, it should be implicit differentiation, but how do I start?

If I let $u = x \cos x$, then I get

$ \begin{eqnarray} \frac{dY}{dX} &=& u \cdot 5^{u - 1} \cdot \frac{du}{dX} \\ &=& x \cdot \cos x \cdot 5^{x \cos x-1} \cdot (\cos x-x \sin x) \end{eqnarray} $

I don't suppose I did it right... The correct answer is

$ \ln5 \cdot 5^{x \cos x}\cdot (\cos x-x \sin x) .$

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    jiewmeng: For x>0 (in real numbers the logarithm isn't defined for $x\le0$ anyway), we can write $x=e^w$ for some $w$, in which case $\large e^{\ln x}=e^{\ln e^w}=e^w=x.$2011-10-17

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Your error is that the derivative of $5^u$ (with respect to $u$) is not $u5^{u-1}$; this isn't $u$ to a power but rather a base to the power of $u$. The usual route is with André's advice: write the function as an exponential, so $5^{x\cos x}=(e^{\ln5})^{x\cos x}=\exp(\ln5\cdot x\cos x)$. Here we use the fact that the natural logarithm and exponential functions are inverse functions of each other, so $\exp\ln a=\ln\exp a=a $.

Now apply the chain rule, and remember the exponential function is its own derivative.

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    @J.M. and anon: My point was just a *mention*...may be then how it works or in reverse order. Once, simplified (and probably knowing *how it works*), I always found it is useful to use the *ready-made rule*;, e.g., to use Binomial Theorem to expand $(a+b)^8$ than multiplying 8 times or in this case using the original definition of derivative as limits :)2011-10-17