I'm in a single-variable calculus course, in which we recently covered logarithmic differentiation. The professor proved it that works when $f(x)>0$, and when $f(x)<0$. I've been trying to find a way to derive that kind of function when $f(x)=0$, but I'm not sure if it's possible, or what. I've thought of this example, that resists all my efforts to differentiate, but seems to be differentiable, (and even appears to have a value of zero).
Find
$$f'\left(\frac{3\pi}{2}\right)\quad \rm where \quad f(x)=(\sin{x} + 1)^x .$$
Is there any way I can find this derivative (if it exists), beyond numerically computing the limit from the definition of the derivative? Or, vice versa, how can I prove that this derivative doesn't exist?
Thanks,
Reggie
Best Answer
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
$$ f'\left({3\pi\over2}\right)=0$$