Find the derivative of a difficult integral

Question

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 $a<k<b$ 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 $f(x)$

Answer 1

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

Answer 2

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.

Answer 3

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 $$ F'(x) = \frac{1}{x} f(y) = \frac{f(\ln x)}{x}\;. $$