# Find the derivative of a difficult integral

This exercise is very difficult for me. Find the derivative of the function: $$\int_{0}^{\ln x}f(t) dt$$

I use this formula: $$\int(b(x)) \cdot b'(x) - \int(a(x)) \cdot a'(x)$$ where this is $$b'(x)$$ the derivative of $$b$$ and this is $$a'(x)$$ the derivative of $$a$$.

And the end my answer is $$\int(\ln(x)) \cdot \frac{1}{x}$$

My question is: Is my answer right?

The problem is that the condition was to use the definition of $$F(x) = \lim_{\varepsilon \rightarrow 0}\frac{1}{\varepsilon }[F(x+\varepsilon) - F(x) ].$$ Also I have to use if the function is continuous in the interval [a,b], then there exists a point $$k$$ for which $$a satisfying: $$\int_{a}^{b}f(x)dx = (b-a)f(k).$$ Finally, I make a boundary transition $$\varepsilon \to 0$$ and use the continuity of $$f(x)$$

• $$\frac{f(\log (x))}{x}$$ Mar 19 at 8:03
• What you're looking for is Leibniz's integral rule : en.wikipedia.org/wiki/… Mar 19 at 8:08
• TL;DR but derivative with respect to what?
– D S
Mar 19 at 10:51
• $\int(\ln(x))\frac1x$ can't be right: there's no $f$ and there's no $dx$ (and if you meant $\int\ln x dx$, then you would want to find that integral instead of leaving it as it is). Mar 20 at 1:35
• Maybe there is a mix-up of $\int$ and $f$, which do look similar. Jul 14 at 13:33

Using the Fundamental Theorem of Calculus and the Chain rule you have: $$\frac{d}{dx}\int_0^{\log x}f(t)dt=f(\log x)\frac{d}{dx}(\log x)=\frac{f(\log x)}{x}$$

Note first of all that you can calculate the derivative using the chain rule. Set $$g(x) = \int_0^x f(t)\ dt$$. Then the function you want to differentiate is $$g(\text{ln}(x))$$. By the chain rule its derivative is $$g'(\text{ln}(x))\frac{1}{x}$$. The fundamental theorem of calculus gives you that $$g'(x) = f(x)$$. Hence the result is $$f(\text{ln}(x))\frac{1}{x}$$.

You can also solve this using the strategy you want to employ. Fix $$x > 0$$ and write down the difference quotient:

$$\frac{1}{h}\Bigl[\int_0^{\text{ln}(x+h)} f(t)\ dt - \int_0^{\text{ln}(x)} f(t)\ dt\Bigr]$$

$$=\frac{1}{h}\int_{\text{ln}(x)}^{\text{ln}(x+h)} f(t)\ dt= \frac{1}{h}(\text{ln}(x+h)-\text{ln}(x))f(c)$$

for some $$c \in (\text{ln}(x),\text{ln}(x+h))$$. If we now let $$h$$ go to $$0$$, we see that $$c$$ converges to $$\text{ln}(x)$$ and the other term converges to the derivative of $$\text{ln}(x)$$ which is $$\frac{1}{x}$$. This agrees with the first result.

Let us start from the definition of derivative: $$F'(x)= \lim \frac{F(x+\Delta x) - F(x)}{\Delta x} = \lim \frac{1}{\Delta x} \left[ \int_0^{\ln(x+\Delta x)} f(t)dt -\int_0^{\ln x} f(t)dt \right]$$ $$= \lim\frac{1}{\Delta x}\int_{\ln x}^{\ln(x+\Delta x)}f(t)dt\;.$$ Introducing a new variable, $$y\equiv\ln x\;,$$ we obtain $$\Delta y\equiv\ln(x+\Delta x) - \ln x\;,\qquad \frac{dy}{dx} = \frac{d\ln x}{dx}=\frac{1}{x}\;. \qquad\qquad\qquad(1)$$ Using the chain rule, $$\lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}\frac{\Delta F}{\Delta y} = \lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x} \;\lim_{\Delta y\rightarrow 0}\frac{\Delta F}{\Delta y}$$, we have: $$F'(x) = \lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}\frac{1}{\Delta y}\int_y^{y+\Delta y} f(t) dt = \lim_{\Delta x\rightarrow 0}\frac{\Delta y}{\Delta x}\;\lim_{\Delta y\rightarrow 0}\frac{1}{\Delta y}\int_y^{y+\Delta y} f(t) dt$$  $$=\frac{dy}{dx}\;\lim\frac{1}{\Delta y}\int_y^{y+\Delta y} f(t) dt \;.$$ From the last formula in your post, it follows that
$$\lim\frac{1}{\Delta y}\int_y^{y+\Delta y} f(t) dt=f(y)\;.$$ Together with (1), this gives us $$F'(x) = \frac{1}{x} f(y) = \frac{f(\ln x)}{x}\;.$$

• starting with the definition of derivative only means you’re going to end up proving the FTC and the chain rule (something which you didn’t prove an airtight proof for)… but anyway it doesn’t seem OP is too hung up about getting things 100% right, so I guess this is a fine answer for OP. Mar 19 at 8:33
• @peek-a-boo Right point. I have now mentioned the chain rule explicitly. Mar 19 at 8:51