Differentiate the integral

Relevant wiki: Differentiation Under the Integral Sign

We can solve this problem in two ways. We can find the expression for $f(x)$ first, and then differentiate it. Another way is to use the general form of Leibniz integral rule

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Method 1 : Finding the expression for $f(x)$

We can treat $x$ as a constant, since we are integrating with respect to $t$ .

$\begin{aligned} f(x) & = \int _{x} ^{x^{2}} \frac{x}{t} \, dt \\ & = \big[ x \ln(t) \big] ^{x^2} _x \\ & = x \ln(x^2) - x \ln (x) \\ & = x \ln \left(\frac{x^2}{x} \right) \\ & = x \ln(x) \end{aligned}$

The integral is equal to $f(x) = x \ln (x)$ . Note that this is defined only for positive $x$ .

We can find its derivative using the product rule . $f'(x) = 1 + \ln x$

When we substitute $x = e^2$ , we get $f'(e^2) = 1 + \ln(e^2) = 1 + 2 = 3$ .

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Method 2: Using the general form of the Leibniz integral rule

If $g(x,t)$ and its partial derivative with respect to $x$ , $g_x(x, t)$ are continuous in $x \leq t \leq x^2$ , and if $a(x), b(x), a'(x), b'(x)$ are all continuous if $x$ belongs to the interval $x_0 \leq x \leq x_1$ ; then for the interval $x_0 \leq x \leq x_1$ , the following holds true:

$\frac{d}{dx} \left( \int_{a(x)} ^{b(x)} g(x, t) \, dt \right) = g(x, b(x)) \cdot b'(x) - g(x, a(x)) \cdot a'(x) + \int _{a(x)} ^{b(x)} \frac{\partial}{\partial x} g(x, t) \, dt$

In our problem, $a(x) = x$ ; $b(x) = x^2$ , $g(x, t) = \frac{x}{t}$ .

$f'(x) = \frac{x}{x^2} \cdot (2x) - \frac{x}{x} \cdot 1 + \int _{x} ^{x^2} \frac{1}{t} \, dt$

On simplification, we obtain $f'(x) = 1 + \ln x$ . Substituting $x = e^2$ yields $f'(e^2) = 3$ $_\square$