Use the Fundamental Theorem of Calculus to find the derivative of $h(x) = \int_{1}^{e^x} \ln (t) \,dt$

The fundamental theorem of calculus states:

If $$f$$ is continuous on $$[a,b]$$, then if $$g(x) = \int_{a}^{x}f(t)\,dt,~ \textrm{then}~g'(x) = f(x)$$.

In Example 4, the chain rule is used because the upper bound, $$x^4$$ needed to be differentiated. My question is, why is that and where is that implied given the theorem stated above?

Similarly, find $$h'(x)$$ of $$h(x) = \int_{1}^{e^x} \ln(t)\,dt$$ The answer is $$h'(x) = xe^x$$, which seems to be in line with Example 4. Per previous examples in the book, one might expect the answer to be $$h'(x) = \ln(e^x)$$ though.

With $$u := e^x$$ the derivative becomes $$\frac{d}{du} \int_1^u \ln(t) dt \frac{du}{dx} = \ln(u) \frac{du}{dx} = \ln(e^x) \frac{d e^x}{dx} = x \cdot e^x.$$

If $$g(x) = \int_{a}^{x}f(t)dt$$, then, by the chain rule, $$\frac{d}{dx}(g(h(x)))=g'(h(x))\cdot h'(x)=f(h(x))\cdot h'(x)$$ where, at the second step, by the FTC, $$g'(h(x))=\left(\frac{d}{dy}\left(\int_{a}^{y}f(t)dt\right)\right)_{y=h(x)}=f(h(x))$$.

So, as regards your example with $$f(x)=\ln(x)$$, $$a=1$$ and $$h(x)=e^x$$, we have that $$\frac{d}{dx}\left( \int_{1}^{e^x} \ln(t)\,dt\right)=\ln(e^x)\cdot (e^x)'=x\cdot e^x.$$ P.S. Note that $$\int_{1}^{e^x} \ln(t)\,dt=\left[\ln(t)t-t\right]_1^{e^x}=xe^x-e^x+1$$ and it is easy to check that $$\frac{d}{dx}\left( xe^x-e^x+1\right)=xe^x$$.

• My confusion is comings from WHY you have to use the chain rule, rather then how the chain rule is used in this problem – Evan Kim Feb 7 at 16:30
• We have to use the chain rule because we have to differentiate a composition of two functions, $g(h(x))$. The FTC tells use how to compute the derivative of $g$. – Robert Z Feb 7 at 16:36

In contrast with the other answers, you could use the other form of the fundamental theorem of calculus which states that (under suitable conditions) if $$F'(t) = f(t)$$ then $$\displaystyle \int_a^b f(t) \, dt = F(b) - F(a).$$ With $$f(t) = \ln t$$ you can take $$F(t) = t \ln t - t$$ so that $$h(x) = \int_1^{e^x} \ln t \, dt = (e^x \ln(e^x) - e^x) - (0 - 1) = x e^x - e^x + 1$$ and $$h'(x) = xe^x + e^x - e^x = xe^x.$$

Do a concrete example to get a feel first: \begin{align} \frac{d}{dx}\left(h(x)\right) &=\frac{d}{dx}\left(\int_{1}^{e^x}\ln{t}\,dt\right)\\ &=\frac{d}{dx}\left(\left[t\ln{t}-t\right]_{1}^{e^x}\right)\\ &=\frac{d}{dx}\left(\left[e^x\ln{e^x}-e^x-(e\ln{1}-1)\right]\right)\\ &=\frac{d}{dx}\left(xe^x-e^x+1\right)\\ &=\frac{d}{dx}(xe^x)-\frac{d}{dx}(e^x)+\frac{d}{dx}(1)\\ &=e^x+xe^x-e^x+0\\ &=xe^x. \end{align}

Then do another one, but this time for a general case:

\begin{align} \frac{d}{dx}\left(\int_{a}^{u(x)}f(t)\,dt\right) &=\frac{d}{dx}\left(\left[F(t)\right]_{a}^{u(x)}\right)\\ &=\frac{d}{dx}\left(F(u(x))-F(a)\right)\\ &=\frac{d}{dx}\left(F(u(x))\right)-\frac{d}{dx}(F(a))\\ &=F'(u(x))-0\\ &=F'(u)\cdot u'(x)\\ &=f(u)\cdot u'(x)\\ &=f(u)\cdot\frac{du}{dx} \end{align}

Where $$F'(x)=f(x)$$.