Calculus Examples

$k e^{- 2 x} - 2 x^{k^{2}}$

Step 1

By the Sum Rule, the derivative of $k e^{- 2 x} - 2 x^{k^{2}}$ with respect to $x$ is $\frac{d}{d x} [k e^{- 2 x}] + \frac{d}{d x} [- 2 x^{k^{2}}]$ .

$\frac{d}{d x} [k e^{- 2 x}] + \frac{d}{d x} [- 2 x^{k^{2}}]$

Step 2

Evaluate $\frac{d}{d x} [k e^{- 2 x}]$ .

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Step 2.1

Since $k$ is constant with respect to $x$ , the derivative of $k e^{- 2 x}$ with respect to $x$ is $k \frac{d}{d x} [e^{- 2 x}]$ .

$k \frac{d}{d x} [e^{- 2 x}] + \frac{d}{d x} [- 2 x^{k^{2}}]$

Step 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 $g (x) = - 2 x$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $- 2 x$ .

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

Step 2.2.2

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

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

Step 2.2.3

Replace all occurrences of $u$ with $- 2 x$ .

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

Step 2.3

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x$ with respect to $x$ is $- 2 \frac{d}{d x} [x]$ .

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

Step 2.4

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

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

Step 2.5

Multiply $- 2$ by $1$ .

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

Step 2.6

Move $- 2$ to the left of $e^{- 2 x}$ .

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

Step 3

Evaluate $\frac{d}{d x} [- 2 x^{k^{2}}]$ .

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Step 3.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x^{k^{2}}$ with respect to $x$ is $- 2 \frac{d}{d x} [x^{k^{2}}]$ .

$k (- 2 e^{- 2 x}) - 2 \frac{d}{d x} [x^{k^{2}}]$

Step 3.2

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

$k (- 2 e^{- 2 x}) - 2 k^{2} x^{k^{2} - 1}$

Step 4

Simplify.

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Step 4.1

Reorder terms.

$- 2 e^{- 2 x} k - 2 x^{k^{2} - 1} k^{2}$

Step 4.2

Reorder factors in $- 2 e^{- 2 x} k - 2 x^{k^{2} - 1} k^{2}$ .

$- 2 k e^{- 2 x} - 2 k^{2} x^{k^{2} - 1}$