Find dy/dx y = square root of e^x+8

$y = \sqrt{e^{x} + 8}$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{e^{x} + 8}$ as ${(e^{x} + 8)}^{\frac{1}{2}}$ .

$y = {(e^{x} + 8)}^{\frac{1}{2}}$

Step 2

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} ({(e^{x} + 8)}^{\frac{1}{2}})$

Step 3

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 4

Differentiate the right side of the equation.

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = e^{x} + 8$ .

To apply the Chain Rule, set $u$ as $e^{x} + 8$ .

$\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [e^{x} + 8]$

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = \frac{1}{2}$ .

$\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [e^{x} + 8]$

Replace all occurrences of $u$ with $e^{x} + 8$ .

$\frac{1}{2} {(e^{x} + 8)}^{\frac{1}{2} - 1} \frac{d}{d x} [e^{x} + 8]$

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{2} {(e^{x} + 8)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [e^{x} + 8]$

Combine $- 1$ and $\frac{2}{2}$ .

$\frac{1}{2} {(e^{x} + 8)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [e^{x} + 8]$

Combine the numerators over the common denominator.

$\frac{1}{2} {(e^{x} + 8)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [e^{x} + 8]$

Simplify the numerator.

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(e^{x} + 8)}^{\frac{1 - 2}{2}} \frac{d}{d x} [e^{x} + 8]$

Subtract $2$ from $1$ .

$\frac{1}{2} {(e^{x} + 8)}^{\frac{- 1}{2}} \frac{d}{d x} [e^{x} + 8]$

Differentiate using the Sum Rule.

Move the negative in front of the fraction.

$\frac{1}{2} {(e^{x} + 8)}^{- \frac{1}{2}} \frac{d}{d x} [e^{x} + 8]$

Combine fractions.

Combine $\frac{1}{2}$ and ${(e^{x} + 8)}^{- \frac{1}{2}}$ .

$\frac{{(e^{x} + 8)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [e^{x} + 8]$

Move ${(e^{x} + 8)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2 {(e^{x} + 8)}^{\frac{1}{2}}} \frac{d}{d x} [e^{x} + 8]$

By the Sum Rule, the derivative of $e^{x} + 8$ with respect to $x$ is $\frac{d}{d x} [e^{x}] + \frac{d}{d x} [8]$ .

$\frac{1}{2 {(e^{x} + 8)}^{\frac{1}{2}}} (\frac{d}{d x} [e^{x}] + \frac{d}{d x} [8])$

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

$\frac{1}{2 {(e^{x} + 8)}^{\frac{1}{2}}} (e^{x} + \frac{d}{d x} [8])$

Differentiate using the Constant Rule.

Since $8$ is constant with respect to $x$ , the derivative of $8$ with respect to $x$ is $0$ .

$\frac{1}{2 {(e^{x} + 8)}^{\frac{1}{2}}} (e^{x} + 0)$

Combine fractions.

Add $e^{x}$ and $0$ .

$\frac{1}{2 {(e^{x} + 8)}^{\frac{1}{2}}} e^{x}$

Combine $\frac{1}{2 {(e^{x} + 8)}^{\frac{1}{2}}}$ and $e^{x}$ .

$\frac{e^{x}}{2 {(e^{x} + 8)}^{\frac{1}{2}}}$

Step 5

Reform the equation by setting the left side equal to the right side.

$y' = \frac{e^{x}}{2 {(e^{x} + 8)}^{\frac{1}{2}}}$

Step 6

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = \frac{e^{x}}{2 {(e^{x} + 8)}^{\frac{1}{2}}}$

Find dy/dx y = square root of e^x+8

Related Questions

Support LCMGCF.com by disabling your adblocker.