Question

Write the composite function in the form \(f\left( {g\left( x \right)} \right)\). (Identify the inner function \(u = g\left( x \right)\) and the outer function \(y = f\left( u \right)\).) Then find the derivative \(\frac{{dy}}{{dx}}\).

5. \(y = {e^{\sqrt x }}\)

Short answer

The inner function and outer function are \(u = g\left( x \right) = \sqrt x \) and \(y = f\left( u \right) = {e^u}\). The derivative of the function is \(\frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{2\sqrt x }}\).

Step-by-step solution

The Chain Rule

The composite function \(F = f \circ g\)is denoted by \(F\left( x \right) = f\left( {g\left( x \right)} \right)\). The differentiation of this function is obtained by using chain rule as shown below:

\(F'\left( x \right) = f'\left( {g\left( x \right)} \right) \cdot g'\left( x \right)\)

By using Leibniz notation,it is defined as:

\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}\)

Identify the inner function and outer function

Consider \(u = g\left( x \right) = \sqrt x \) as the inner function.

Let \(y = f\left( u \right) = {e^u}\) be the outer function.

Differentiate the function

Use Leibniz notation to differentiate the function as shown below:

\(\begin{aligned}\frac{{dy}}{{dx}} &= \frac{{dy}}{{du}} \cdot \frac{{du}}{{dx}}\\ &= \frac{d}{{du}}\left( {{e^u}} \right) \cdot \frac{d}{{dx}}\left( {\sqrt x } \right)\\ &= \frac{d}{{du}}\left( {{e^u}} \right) \cdot \frac{d}{{dx}}\left( {{x^{\frac{1}{2}}}} \right)\\ &= {e^u} \cdot \frac{1}{2}{x^{ - \frac{1}{2}}}\\ &= {e^{\sqrt x }} \cdot \frac{1}{{2\sqrt x }}\\ &= \frac{{{e^{\sqrt x }}}}{{2\sqrt x }}\end{aligned}\)

Thus, the derivative of the function is \(\frac{{dy}}{{dx}} = \frac{{{e^{\sqrt x }}}}{{2\sqrt x }}\).