Algebra Examples

$y = 7 e^{- x} + e^{4 x}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (7 e^{- x} + e^{4 x})$

Step 2

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

$y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $7 e^{- x} + e^{4 x}$ with respect to $x$ is $\frac{d}{d x} [7 e^{- x}] + \frac{d}{d x} [e^{4 x}]$ .

$\frac{d}{d x} [7 e^{- x}] + \frac{d}{d x} [e^{4 x}]$

Step 3.2

Evaluate $\frac{d}{d x} [7 e^{- x}]$ .

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Step 3.2.1

Since $7$ is constant with respect to $x$ , the derivative of $7 e^{- x}$ with respect to $x$ is $7 \frac{d}{d x} [e^{- x}]$ .

$7 \frac{d}{d x} [e^{- x}] + \frac{d}{d x} [e^{4 x}]$

Step 3.2.2

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

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Step 3.2.2.1

To apply the Chain Rule, set $u_{1}$ as $- x$ .

$7 (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [- x]) + \frac{d}{d x} [e^{4 x}]$

Step 3.2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$7 (e^{u_{1}} \frac{d}{d x} [- x]) + \frac{d}{d x} [e^{4 x}]$

Step 3.2.2.3

Replace all occurrences of $u_{1}$ with $- x$ .

$7 (e^{- x} \frac{d}{d x} [- x]) + \frac{d}{d x} [e^{4 x}]$

Step 3.2.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$7 (e^{- x} (- \frac{d}{d x} [x])) + \frac{d}{d x} [e^{4 x}]$

Step 3.2.4

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

$7 (e^{- x} (- 1 \cdot 1)) + \frac{d}{d x} [e^{4 x}]$

Step 3.2.5

Multiply $- 1$ by $1$ .

$7 (e^{- x} \cdot - 1) + \frac{d}{d x} [e^{4 x}]$

Step 3.2.6

Move $- 1$ to the left of $e^{- x}$ .

$7 (- 1 \cdot e^{- x}) + \frac{d}{d x} [e^{4 x}]$

Step 3.2.7

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$7 (- e^{- x}) + \frac{d}{d x} [e^{4 x}]$

Step 3.2.8

Multiply $- 1$ by $7$ .

$- 7 e^{- x} + \frac{d}{d x} [e^{4 x}]$

Step 3.3

Evaluate $\frac{d}{d x} [e^{4 x}]$ .

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Step 3.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 4 x$ .

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Step 3.3.1.1

To apply the Chain Rule, set $u_{2}$ as $4 x$ .

$- 7 e^{- x} + \frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [4 x]$

Step 3.3.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$- 7 e^{- x} + e^{u_{2}} \frac{d}{d x} [4 x]$

Step 3.3.1.3

Replace all occurrences of $u_{2}$ with $4 x$ .

$- 7 e^{- x} + e^{4 x} \frac{d}{d x} [4 x]$

Step 3.3.2

Since $4$ is constant with respect to $x$ , the derivative of $4 x$ with respect to $x$ is $4 \frac{d}{d x} [x]$ .

$- 7 e^{- x} + e^{4 x} (4 \frac{d}{d x} [x])$

Step 3.3.3

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

$- 7 e^{- x} + e^{4 x} (4 \cdot 1)$

Step 3.3.4

Multiply $4$ by $1$ .

$- 7 e^{- x} + e^{4 x} \cdot 4$

Step 3.3.5

Move $4$ to the left of $e^{4 x}$ .

$- 7 e^{- x} + 4 e^{4 x}$

Step 3.4

Reorder terms.

$4 e^{4 x} - 7 e^{- x}$

Step 4

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

$y' = 4 e^{4 x} - 7 e^{- x}$

Step 5

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

$\frac{d y}{d x} = 4 e^{4 x} - 7 e^{- x}$