Question

Find the derivative of the function.

\(y = \int_{\sqrt x }^x {\frac{{{e^t}}}{t}} dt\)

Short answer

The value of the derivative for the integral \(y = \int_{\sqrt x }^x {\frac{{{e^t}}}{t}} dt\) is \(\frac{{{e^x}}}{x} - \frac{{{e^{\sqrt x }}}}{{2x}}\).

Step-by-step solution

Function definition

In mathematics, a function is an expression, rule, or law that describes the connection between one variable and another variable. In mathematics, functions are everywhere, and they're crucial for articulating physical links in the sciences.

Applying the Fundamental Theorem of Calculus

The fundamental theorem of calculus is a theorem that connects the concepts of differentiating (calculating the gradient) and integrating (calculating the slope) (calculating the area under the curve).

Suppose\(f\)is continuous on \((a,b)\), if \(g(x) = \int\limits_a^x f (t)dt\;\) then \({g^'}(x) = f(x)\).

\(y = \int_{\sqrt x }^x {\frac{{{e^t}}}{t}} dt\)

Evaluating the integral

Split the integral using \(\int_a^c f (x)dx = \int_a^b f (x)dx + \int_b^c f (x)dx\), here \(a = 0,b = \sqrt x \) and \(c = x\).

\(\begin{aligned}{c}\int_0^x {\frac{{{e^t}}}{t}} dt &= \int_0^{\sqrt x } {\frac{{{e^l}}}{t}} dt + \int_{\sqrt x }^x {\frac{{{e^t}}}{t}} dt\\\int_{\sqrt x }^x {\frac{{{e^t}}}{t}} dt &= \int_0^x {\frac{{{e^t}}}{t}} dt - \int_0^{\sqrt x } {\frac{{{e^t}}}{t}} dt\end{aligned}\)

Taking derivative on both the sides –

\(\begin{aligned}{c}\frac{d}{{dx}}\int_{\sqrt x }^x {\frac{{{e^t}}}{t}} dt &= \frac{d}{{dx}}\left( {\int_0^x {\frac{{{e^t}}}{t}} dt - \int_0^{\sqrt x } {\frac{{{e^t}}}{t}} dt} \right)\\\frac{d}{{dx}}\int_{\sqrt x }^x {\frac{{{e^t}}}{t}} dt &= \frac{d}{{dx}}\int_0^x {\frac{{{e^t}}}{t}} dt - \frac{d}{{dx}}\int_0^{\sqrt x } {\frac{{{e^t}}}{t}} dt.........(1)\end{aligned}\)

Simplifying and deducing the derivative

Solving RHS separately –

\({F^\prime }(x) = \frac{d}{{dx}}\int_0^x {\frac{{{e^t}}}{t}} dt\)

Here,\(f(t) = \frac{{{e^t}}}{t}\)considering the fundamental theorem of calculus\({F^\prime }(x) = f(x)\)-

\({F^\prime }(x) = \frac{{{e^x}}}{x}\)

\(G(x) = \int_0^{\sqrt x } {\frac{{{e^t}}}{t}} dt = H(\sqrt x ) - H(0)\;\) \(H(t)\)is the antiderivative of \(h(t) = \frac{{{e^t}}}{t}\).

\(\begin{aligned}{c}{G^\prime }(x) &= \frac{d}{{dx}}\int_0^{\sqrt x } {\frac{{{e^t}}}{t}} dt = {H^\prime }(\sqrt x ) \cdot \frac{d}{{dx}}\sqrt x - 0\\{G^\prime }(x) &= h(\sqrt x ) \cdot \frac{1}{{2\sqrt x }}\\{G^\prime }(x) &= \frac{{{e^{\sqrt x }}}}{{\sqrt x }} \cdot \frac{1}{{2\sqrt x }} = \frac{{{e^{\sqrt x }}}}{{2x}}\end{aligned}\)

Now plugging in both \({F^\prime }(x)\)and \({G^\prime }(x)\)in equation \((1)\)-

\(\frac{d}{{dx}}\int_{\sqrt x }^x {\frac{{{e^t}}}{t}} dt = \frac{{{e^x}}}{x} - \frac{{{e^{\sqrt x }}}}{{2x}}\)

Therefore, the derivative is obtained as \(\frac{{{e^x}}}{x} - \frac{{{e^{\sqrt x }}}}{{2x}}\).