Calculus Examples

$f (x) = (e^{2.5} + \frac{1}{e^{- x}})$

Step 1

Differentiate.

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Step 1.1

By the Sum Rule, the derivative of $e^{2.5} + \frac{1}{e^{- x}}$ with respect to $x$ is $\frac{d}{d x} [e^{2.5}] + \frac{d}{d x} [\frac{1}{e^{- x}}]$ .

$\frac{d}{d x} [e^{2.5}] + \frac{d}{d x} [\frac{1}{e^{- x}}]$

Step 1.2

Since $e^{2.5}$ is constant with respect to $x$ , the derivative of $e^{2.5}$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [\frac{1}{e^{- x}}]$

Step 1.3

Simplify the expression.

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Step 1.3.1

Add $0$ and $\frac{d}{d x} [\frac{1}{e^{- x}}]$ .

$\frac{d}{d x} [\frac{1}{e^{- x}}]$

Step 1.3.2

Rewrite $\frac{1}{e^{- x}}$ as ${(e^{- x})}^{- 1}$ .

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

Step 1.3.3

Multiply the exponents in ${(e^{- x})}^{- 1}$ .

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Step 1.3.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{d}{d x} [e^{- x \cdot - 1}]$

Step 1.3.3.2

Multiply $- x \cdot - 1$ .

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Step 1.3.3.2.1

Multiply $- 1$ by $- 1$ .

$\frac{d}{d x} [e^{1 x}]$

Step 1.3.3.2.2

Multiply $x$ by $1$ .

$\frac{d}{d x} [e^{x}]$

Step 2

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

$e^{x}$