How to differentiate $F\left(x\right)=\int _1^x\frac{x}{\sqrt{t}}dt$

Question

How to differentiate $F\left(x\right)=\int _1^x\frac{x}{\sqrt{t}}dt$?

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Since we are differentiating with respect to $t$, I can treat the $x$ as constant.

So,

$$F\left(x\right)=x\int _1^x\frac{1}{\sqrt{t}}dt$$

What would $F'(x)$ be?

Since we are treating $x$ as constant, I think it would be:

$$F'(x) = x \frac{1}{\sqrt{x}}$$ but according to desmos this is not the right answer.

The orange line should match with the green one. What am I doing wrong?

Answer 1

You should apply product rule:

$$F'(x)=x'\int_1^x\frac1{\sqrt t}dt+x\left(\int_1^x\frac1{\sqrt t}dt\right)'$$

and we know that $\left(\int_1^x\frac1{\sqrt t}dt\right)'=\frac1{\sqrt x}$.

On a side note, the integral isn't hard to evaluate:

$$\left.F(x)=x\int_1^xt^{-1/2}\ dt=x(2\sqrt t)\right|_1^x$$

Answer 2

You could even evaluate directly the integral:

$$\left.\int_1^x\frac{dt}{\sqrt t}=2\sqrt t\right|_1^x=2\sqrt x-2\implies$$

$$f(x)=2x\sqrt x-2x\implies f'(x)=3\sqrt x-2$$

Answer 3

See you have to applt Leibnitz rule for differentiation under integral sign.

So $F^{'}(x)= [\frac{d}{dx}(x)].(\frac{x}{\sqrt{x}}) - [\frac{d}{dx}(1)]\int_{1}^{x} \frac{1}{\sqrt{1} }dt + \int_{1}^{x}\frac{d}{dx}[\frac{x}{\sqrt{t} }]dt$ giving $F^{'}(x)= 3\sqrt{x} - 2$