Computing the derivative of $H(z)= \int_{e^z}^{\cos z} \ln(w^3) ~dw$ I am stuck on this problem:

Find the derivative of the function: $$H(z)= \int_{e^z}^{\cos z} \ln(w^3) ~dw.$$

 A: We will use a (very) modified brute force approach.  Let $f(w)=\ln(w^3)$. The idea is to find an antiderivative $F(w)$ of $f(w)$. Then $H(z)=F(\cos z)-F(e^z)$. Once we have found $F(w)$, we will have an expression for $H(z)$, and differentiation should be easy.
But we don't need to find $F(w)$!  We have 
$$H(z)=F(\cos z)-F(e^z).$$
Differentiate, using the Chain Rule, and recalling that though we never calculated $F(w)$, we know that $F'(w)=f(w)$. We get
$$H'(z)=(-\sin z)f(\cos z)-(e^z)f(e^z).$$
But $f(w)=\ln(w^3)=3\ln w$. Thus 
$$H'(z)=(-\sin z)(3)(\ln(\cos z)) -(e^z)(3)(z).$$
The answer looks nicer if we write it  as $-3\left[\sin z\,\ln(\cos z)+ze^z\right]$.
Comment: In this case, finding $F(w)$, an antiderivative of $\ln(w^3)$, is easy, particularly if we notice that $\ln(w^3)=3\ln w$.  However, the technique we used works for "any" $f(w)$, even if we don't know an antiderivative of $f(w)$.
I went through the full reasoning, but one use the reasoning to obtain a general formula, and after that just use the formula.
Let $f(w)$, $a(z)$, and $b(z)$ be nice functions, and let
$$H(z)=\int_{a(z)}^{b(z)}f(w)\,dw.$$
Then 
$$H'(z)=b'(z)f(b(z))-a'(z)f(a(z)).$$
A: Hint:
If $g(x)$ and $h(x)$ are differentiable and $f(u)$ is a continuous function between $g(x)$ and $h(x)$, then $$\frac{d}{dx}\int_{g(x)}^{h(x)} f(u)~du = f(g(x))\cdot g'(x)-f(h(x))\cdot h'(x)$$
A: By the Second Fundamental Theorem of Calculus, if $f(u)$ is continuous we have:
$$\frac{d}{dx} \int_a^x f(u)\,du = f(x).$$
Using the Chain Rule, if $h(x)$ is any function, we have:
$$\frac{d}{dx} \int_a^{h(x)} f(u)\,du = f(h(x))h'(x).$$
Using this and properties of the integral, we have:
$$\begin{align*}
\frac{d}{dz}H(z) &= \frac{d}{dz}\int_{e^z}^{\cos z} \ln(w^3)\,dw\\
&=\frac{d}{dz}\left(\int_{e^z}^1\ln(w^3)\,dw +\int_1^{\cos z}\ln(w^3)\,dw\right)\\
&=\frac{d}{dz}\left(\int_1^{\cos z}\ln(w^3)\,dw - \int_1^{e^z}\ln(w^3)\,dw\right)\\
&=\frac{d}{dz}\int_1^{\cos z}\ln(w^3)\,dw - \frac{d}{dz}\int_1^{e^z}\ln(w^3)\,dw\\
&= \ln(\cos^3(z))(\cos z)' - \ln((e^z)^3)(e^z)'\\
&= -3\sin z\ln(\cos z) - 3ze^z.
\end{align*}$$
A: A brute-force approach:


*

*Use change of variable: $x = w^3$

*Use integration by parts on the resulting integral to integrate the function.

*Find the derivative the usual way using chain rule/product rules etc.
