Difficult Derivative?
Mia Lopez 
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
$\endgroup$ 18 Answers
$\begingroup$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
$$ f'\left({3\pi\over2}\right)=0$$
$\endgroup$ $\begingroup$$$\log{f(x)} = x \log{(\sin{x}+1)}$$
$$\frac{d}{dx} \log{f(x)} = \log{(\sin{x}+1)} + \frac{x \cos{x}}{\sin{x}+1} = \frac{(\sin{x}+1)\log{(\sin{x}+1)}+x \cos{x}}{\sin{x}+1} $$
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, $f'(3 \pi/2)=0$.
$\endgroup$ 4 $\begingroup$$f(x) = (sin(x)+1)^x = (e^{ln(sin(x)+1)})^x = e^{x ln(sin(x)+1)}$.
Then $f'(x) = e^{x ln(sin(x)+1)} * \frac{d}{dx} [x ln(sin(x)+1)]$
by the rules for differentiating exponentials, and the chain rule.
Now $\frac{d}{dx} [x ln(sin(x)+1)] = ln(sin(x)+1) + x * \frac{1}{sin(x)+1} * cos(x)$
by a combination of the product rule and chain rule.
Now simply insert the value of $\frac{3\pi}{2}$ and see what happens (I don't have time to do this at the moment).
$\endgroup$ $\begingroup$Graph of $f(x)$:
Graph of $f'(x)$:
See both the above graphs. $f(x)$ is actually not differentiable at $x= 1.5π$. The graph of $f'(x)$ at $x = 1.5π$ is a vertical asymptote. The function's second differential may say that it is increasing/decreasing at $x= 1.5π$ but the first derivative doesn't exist.
$\endgroup$ 2 $\begingroup$Most people are unaware that there is actually a rule that governs $q^r$, assuming that the domain is valid. It's actually a combination of the power rule and the exponent rule:
$$ d(q^r) = \ln(q)\,q^r\,dr + r\,q^{r-1}dr $$
Therefore, for $(\sin(x) + 1)^x$, you could make $q = \sin(x) + 1$ and $dq = \cos(x)\,dx$. This would give you:
$$d\left((\sin(x) + 1)^x\right) = \ln(\sin(x) + 1)\,(\sin(x) + 1)^x\,dx + x\,(\sin(x) + 1)^{x - 1}\,\cos(x)\, dx$$
Therefore, to get $\frac{df}{dx}$ just divide by $dx$:
$$ f'(x) = \ln(\sin(x) + 1)\,(\sin(x) + 1)^x + x\,(\sin(x) + 1)^{x - 1}\,\cos(x) $$
$\endgroup$ $\begingroup$Not completely sure what you're looking for here, but a simple way of solving this is the chain rule.
$f(x)=(\sin{x} +1)^x$
First you have to realize that you're dealing with a chain in the form of $a^x$. Since the derivative of $a^x$ is $\ln(a)\cdot a^x$.
Set $\sin(x)+1$ equal to $u$Then your function looks like this
$f(x)=u^x$
In the chain rule, you take the derivative and write ignore the $u$ and then multiply it by the derivative of the $u$.
We will take the derivative of $u^x$ then multiply it by the derivative of $u$ Shown here
$f'(x)=\ln(u)\cdot(u^x)\cdot \frac{du}{dx}$
Plug in the blop for u
$f'(x)= \ln(\sin{x} +1)\cdot(\sin{x} +1)^x \cdot\frac{d}{dx} (\sin{x} +1)$
Then take the derivative of the $\sin{x} + 1$ which is $\cos{x}$ because the derivative of a constant is 0
$f(x)= \ln(\sin{x} +1)\cdot(\sin{x} +1)^x \cdot \cos{x}$
Then plug in $\frac{3\pi}{2}$
$f'(3π/2)= \ln(\sin\frac{3\pi}{2} +1)\cdot(\sin{\frac{3\pi}{2}})^{(\frac{3\pi}{2})} \cdot \cos\frac{3\pi}{2}$
Since $\cos\frac{3\pi}{2} = 0$ then $f'(\frac{3\pi}{2}) = 0$ because anything multiplied by 0 becomes 0. You didn't have to use the rules of logarithms to accomplish this, unless your teacher specifically asks you to...
$\endgroup$ 2 $\begingroup$One interesting way to go about this is simply utilizing the Chain Rule. As our function is defined as f(x)= (sin(x)+1) raised to the "x-th" Power, we can define our "w" and "u" values in order to apply the Chain Rule.
**W= p (arbitrary variable) raised to the "x-th"
W'= x(p) raised to to the x-1
U= sin(x)+1
U'= cos(x)**
When applying the chain rule we get this result:
[x(sin(x)+1)) raised to the x-1] * cos(x)
As we are evaluating f'(3pi/2), we can substitute 3pi/2 in for the x values of the equation.
As a result we would get:
3pi/2(sin(3pi/2)+1)) raised to the (3pi/2)-1 power * cos(3pi/2)
Simplifying this we see:
[3pi/2(-1+1) raised to the (3pi/2)-1 power] * 0
0 raised to any power other than 0 is 0, multiplied by 0 is zero!
0 raised to the (3pi/2) is 0, multiplied by 0 is still zero
The chain rule is a very interesting way to looking into this!
$\endgroup$ $\begingroup$You can find the first derivative algebraically using the product and chain rules.
I rewrote the function as $y=e$ to the power of $\ln \sin x +1$. Then I rewrote this as e to the power of u. I used the chain rule to find $\frac{dy}{du}$, which was the original function (as the derivative of e to the x is just e to the x). The derivite of the inner function $\ln \sin x+1$ required use of the product and chain rules. We can find the derivitive of the overall function to be $\ln \sin x+1$ times $\cos x \csc x +2 \cos x$. Then we can substitute in for $x$. Notice how $e$ to the power of $ln 0$ is just $0$.
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