Difficult Derivative?

Solution 1:

Oddly enough, this is a case where you actually do well to revert to the definition of the derivative:

$$f'\left({3\pi\over2}\right)=\lim_{h\rightarrow0}{f\left({3\pi\over2}+h\right)-f\left({3\pi\over2}\right)\over h}$$

You don't have to do any numerical computation. Note that $\sin({3\pi\over2}+h)=-\cos h$, so for $f(x)=(1+\sin x)^x$ the definition becomes

$$f'\left({3\pi\over2}\right)=\lim_{h\rightarrow0}{(1-\cos h)^{{3\pi\over2}+h}\over h}$$

since $f({3\pi\over2})$ is clearly $0$. But for (small) nonzero $h$, we have $$0\lt|1-\cos h|^{{3\pi\over2}+h}\lt|1-\cos h|$$since $|1-\cos h|\lt1$ and $3\pi/2\gt1$, from which, using the definition of the derivative for the cosine function and the fact that $(\cos x)'=-\sin x$, it follows that

$$0\le \lim_{h\rightarrow0}\left|{(1-\cos h)^{{3\pi\over2}+h}\over h}\right|\le\left|\lim_{h\rightarrow0}{1-\cos h\over h}\right|=|\sin0|=0$$

hence

$$ f'\left({3\pi\over2}\right)=0$$

Solution 2:

$$\log{f(x)} = x \log{(\sin{x}+1)}$$

$$\frac{d}{dx} \log{f(x)} = \log{(\sin{x}+1)} + \frac{x \cos{x}}{\sin{x}+1} = \frac{(\sin{x}+1)\log{(\sin{x}+1)}+x \cos{x}}{\sin{x}+1} $$

Now,

$$\frac{d}{dx} \log{f(x)} = \frac{f'(x)}{f(x)}$$

so that

$$f'(x) = (\sin{x}+1)^{x} \log{(\sin{x}+1)}+ (\sin{x}+1)^{x-1} x \cos{x} $$

Note that this has been mere manipulation without any regard to where $f'$ may exist, etc. However, as you are interested in the value at $x=3 \pi/2$, $x \gt 0$, so that the factor $(\sin{x}+1)^{x}$ goes to zero faster than $\log{(\sin{x}+1)}$ blows up. Therefore, $f'(3 \pi/2)=0$.