Question

Use the Second Fundamental Theorem of Calculus, if needed, to calculate each of the derivatives given below.

$\frac{d}{dx}{\int }_x^{\sqrt{x}} \frac{e^x}{x}dt$

Short answer

Ans: $\frac{d}{dx}{\int }_x^{\sqrt{x}} \frac{e^x}{x}dt=\frac{\sqrt{x}}{2}\frac{e^x}{x}-\frac{e^x}{x}.$

Step-by-step solution

Step 1. given information.

given expression, $\frac{d}{dx}{\int }_x^{\sqrt{x}} \frac{e^x}{x}dt$

Step 2. The objective is to calculate the derivative.  

The derivative can be written as,

$\begin{array}{r}\frac{d}{dx}\left({\int }_x^0 \frac{e^x}{x}dt+{\int }_0^{\sqrt{x}} \frac{e^x}{x}dt\right)\\ =\frac{d}{dx}\left(-{\int }_0^x \frac{e^x}{x}dt+{\int }_0^{\sqrt{x}} \frac{e^x}{x}dt\right)\end{array}$

Now, if $f$is continuous on $[a,b]$ then for all $x\in [a,b]$,

$\frac{d}{dx}{\int }_a^{u\left(x\right)} f\left(t\right)dt=f\left(u\right(x\left)\right)u^{\mathrm{\prime }}\left(x\right)$

Step 3. The derivate expression can be written as, 

$\begin{array}{r}\frac{d}{dx}\left(-{\int }_0^x \frac{e^x}{x}dt+{\int }_0^{\sqrt{x}} \frac{e^x}{x}dt\right)\\ =-\frac{e^x}{x}+\frac{1}{2}\sqrt{x}\frac{e^x}{x}\\ =\frac{\sqrt{x}}{2}\frac{e^x}{x}-\frac{e^x}{x}\end{array}$

Therefore, the answer is $\frac{\sqrt{x}}{2}\frac{e^x}{x}-\frac{e^x}{x}$.