Chapter 3.4, Problem 5E

To determine

To find: The composite function in the form $f\left(g\left(x\right)\right)$ and obtain the derivative of y.

Answer to Problem 5E

The inner function is $u=\sqrt{x}$ and the outer function is $f\left(u\right)=e^u$.

The derivative of y is $\underset{\_}{y^{\prime }\left(x\right)=\frac{e^{\sqrt{x}}}{2\sqrt{x}}}$.

Explanation of Solution

Given:

The function is $y=e^{\sqrt{x}}$.

Formula used:

The Chain Rule:

If h is differentiable at x and g is differentiable at $h\left(x\right)$, then the composite function $F=g\circ h$ defined by $F\left(x\right)=g\left(h\left(x\right)\right)$ is differentiable at x and $F^{\prime }$ is given by the product

$F^{\prime }\left(x\right)=g^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)$ (1)

Power Rule:

If n is positive integer, then $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ (2)

Calculation:

Let the inner function be $u=g\left(x\right)$ and the outer function be $y=f\left(u\right)$.

Then, $g\left(x\right)=\sqrt{x}$ and $f\left(u\right)=e^u$. That is,

$\begin{array}{c}y=e^{\sqrt{x}}\\ =f\left(\sqrt{x}\right)\\ =f\left(g\left(x\right)\right)\end{array}$

Therefore, $y=f\left(g\left(x\right)\right)$.

Hence, the inner function is $u=\sqrt{x}$ and the outer function is $f\left(u\right)=e^u$.

Thus, the required form of composite function is $$.

Obtain the derivative of y is

Let $h\left(x\right)=\sqrt{x}$ and $g\left(u\right)=e^u\text{ where }u=h\left(x\right)$

Apply the chain rule as shown in equation (1),

$y^{\prime }\left(x\right)=g^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)$ (3)

The derivative of $g^{\prime }\left(h\left(x\right)\right)$ is computed as follows,

$\begin{array}{c}{g}^{\prime }\left(h\left(x\right)\right)=g^{\prime }\left(u\right)\\ =\frac{d}{du}\left(g\left(u\right)\right)\\ =\frac{d}{du}{e}^u\\ =e^u\end{array}$

Substitute $u=\sqrt{x}$ in above equation,

$g^{\prime }\left(h\left(x\right)\right)=e^{\sqrt{x}}$

Thus, the derivative $g^{\prime }\left(h\left(x\right)\right)$ is $g^{\prime }\left(h\left(x\right)\right)=e^{\sqrt{x}}$.

The derivative of $h\left(x\right)$ is computed as follows,

$\begin{array}{c}{h}^{\prime }\left(x\right)=\frac{d}{dx}\left(\sqrt{x}\right)\\ =\frac{d}{dx}\left(x^{\frac{1}{2}}\right)\end{array}$

Apply the power rule as shown in equation (2),

$\begin{array}{c}{h}^{\prime }\left(x\right)=\frac{1}{2}{x}^{\frac{1}{2}-1}\\ =\frac{1}{2}{x}^{\frac{1-2}{2}}\\ =\frac{1}{2}{x}^{\frac{-1}{2}}\\ =\frac{1}{2\sqrt{x}}\end{array}$

Thus, the derivative of $h\left(x\right)$ is $h^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}$.

Substitute $e^{\sqrt{x}}$ for $g^{\prime }\left(h\left(x\right)\right)$ and $\frac{1}{2\sqrt{x}}$ for $h^{\prime }\left(x\right)$ in equation (3),

$\begin{array}{c}{g}^{\prime }\left(h\left(x\right)\right)\cdot h^{\prime }\left(x\right)=e^{\sqrt{x}}\left(\frac{1}{2\sqrt{x}}\right)\\ =\frac{e^{\sqrt{x}}}{2\sqrt{x}}\end{array}$

Therefore, The derivative of $y=e^{\sqrt{x}}$ is $\underset{\_}{y^{\prime }\left(x\right)=\frac{e^{\sqrt{x}}}{2\sqrt{x}}}$.