Question

Differentiate the function.

23.\(h\left( x \right) = {e^{{x^2} + \ln x}}\)

Short answer

The derivative of the function is \(h'\left( x \right) = {e^{{x^2}}}\left( {2{x^2} + 1} \right)\).

Step-by-step solution

Derivative of logarithmic functions

The derivative of a logarithmicfunctionis shown below:

1. \(\frac{d}{{dx}}\left( {{{\log }_b}x} \right) = \frac{1}{{x\ln b}}\)
2. \(\frac{d}{{dx}}\left( {\ln x} \right) = \frac{1}{x}\)
3. \(\frac{d}{{dx}}\left( {\ln u} \right) = \frac{1}{u}\frac{{du}}{{dx}}\) or \(\frac{d}{{dx}}\left( {\ln g\left( x \right)} \right) = \frac{{g'\left( x \right)}}{{g\left( x \right)}}\)

Differentiate the function

Simplify the function as shown below:

\(\begin{aligned}h\left( x \right)&= {e^{{x^2} + \ln x}}\\&= {e^{{x^2}}} \cdot {e^{\ln x}}\\&= {e^{{x^2}}} \cdot x\\&= x{e^{{x^2}}}\end{aligned}\)

Use the product rule and the chain rule to differentiate the function as shown below:

\(\begin{aligned}{c}h'\left( x \right)&= \frac{d}{{dx}}\left( {x{e^{{x^2}}}} \right)\\&= x \cdot \frac{d}{{dx}}\left( {{e^{{x^2}}}} \right) + {e^{{x^2}}}\frac{d}{{dx}}\left( x \right)\\&= x{e^{{x^2}}}\frac{d}{{dx}}\left( {{x^2}} \right) + {e^{{x^2}}} \cdot \left( 1 \right)\\&= x{e^{{x^2}}} \cdot \left( {2x} \right) + {e^{{x^2}}}\\&= {e^{{x^2}}}\left( {2{x^2} + 1} \right)\end{aligned}\)

Thus, the derivative of the function is \(h'\left( x \right) = {e^{{x^2}}}\left( {2{x^2} + 1} \right)\).