Calculus Examples

$f (x) = - 2 e^{- 5 x} + 15 x^{0.2} - 4$

Step 1

By the Sum Rule, the derivative of $- 2 e^{- 5 x} + 15 x^{0.2} - 4$ with respect to $x$ is $\frac{d}{d x} [- 2 e^{- 5 x}] + \frac{d}{d x} [15 x^{0.2}] + \frac{d}{d x} [- 4]$ .

$\frac{d}{d x} [- 2 e^{- 5 x}] + \frac{d}{d x} [15 x^{0.2}] + \frac{d}{d x} [- 4]$

Step 2

Evaluate $\frac{d}{d x} [- 2 e^{- 5 x}]$ .

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Step 2.1

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 e^{- 5 x}$ with respect to $x$ is $- 2 \frac{d}{d x} [e^{- 5 x}]$ .

$- 2 \frac{d}{d x} [e^{- 5 x}] + \frac{d}{d x} [15 x^{0.2}] + \frac{d}{d x} [- 4]$

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 $g (x) = - 5 x$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $- 5 x$ .

$- 2 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- 5 x]) + \frac{d}{d x} [15 x^{0.2}] + \frac{d}{d x} [- 4]$

Step 2.2.2

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

$- 2 (e^{u} \frac{d}{d x} [- 5 x]) + \frac{d}{d x} [15 x^{0.2}] + \frac{d}{d x} [- 4]$

Step 2.2.3

Replace all occurrences of $u$ with $- 5 x$ .

$- 2 (e^{- 5 x} \frac{d}{d x} [- 5 x]) + \frac{d}{d x} [15 x^{0.2}] + \frac{d}{d x} [- 4]$

Step 2.3

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

$- 2 (e^{- 5 x} (- 5 \frac{d}{d x} [x])) + \frac{d}{d x} [15 x^{0.2}] + \frac{d}{d x} [- 4]$

Step 2.4

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

$- 2 (e^{- 5 x} (- 5 \cdot 1)) + \frac{d}{d x} [15 x^{0.2}] + \frac{d}{d x} [- 4]$

Step 2.5

Multiply $- 5$ by $1$ .

$- 2 (e^{- 5 x} \cdot - 5) + \frac{d}{d x} [15 x^{0.2}] + \frac{d}{d x} [- 4]$

Step 2.6

Move $- 5$ to the left of $e^{- 5 x}$ .

$- 2 (- 5 \cdot e^{- 5 x}) + \frac{d}{d x} [15 x^{0.2}] + \frac{d}{d x} [- 4]$

Step 2.7

Multiply $- 5$ by $- 2$ .

$10 e^{- 5 x} + \frac{d}{d x} [15 x^{0.2}] + \frac{d}{d x} [- 4]$

Step 3

Evaluate $\frac{d}{d x} [15 x^{0.2}]$ .

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Step 3.1

Since $15$ is constant with respect to $x$ , the derivative of $15 x^{0.2}$ with respect to $x$ is $15 \frac{d}{d x} [x^{0.2}]$ .

$10 e^{- 5 x} + 15 \frac{d}{d x} [x^{0.2}] + \frac{d}{d x} [- 4]$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 0.2$ .

$10 e^{- 5 x} + 15 (0.2 x^{- 0.8}) + \frac{d}{d x} [- 4]$

Step 3.3

Multiply $0.2$ by $15$ .

$10 e^{- 5 x} + 3 x^{- 0.8} + \frac{d}{d x} [- 4]$

Step 4

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$10 e^{- 5 x} + 3 x^{- 0.8} + 0$

Step 5

Simplify.

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Step 5.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$10 e^{- 5 x} + 3 \frac{1}{x^{0.8}} + 0$

Step 5.2

Combine terms.

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Step 5.2.1

Combine $3$ and $\frac{1}{x^{0.8}}$ .

$10 e^{- 5 x} + \frac{3}{x^{0.8}} + 0$

Step 5.2.2

Add $10 e^{- 5 x} + \frac{3}{x^{0.8}}$ and $0$ .

$10 e^{- 5 x} + \frac{3}{x^{0.8}}$