Calculus Examples

$y = 5 (1 - e^{- 2 x})$

Step 1

Write $y = 5 (1 - e^{- 2 x})$ as a function.

$f (x) = 5 (1 - e^{- 2 x})$

Step 2

Find the first derivative of the function.

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Step 2.1

Differentiate.

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Step 2.1.1

Since $5$ is constant with respect to $x$ , the derivative of $5 (1 - e^{- 2 x})$ with respect to $x$ is $5 \frac{d}{d x} [1 - e^{- 2 x}]$ .

$5 \frac{d}{d x} [1 - e^{- 2 x}]$

Step 2.1.2

By the Sum Rule, the derivative of $1 - e^{- 2 x}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- e^{- 2 x}]$ .

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

Step 2.1.3

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$5 (0 + \frac{d}{d x} [- e^{- 2 x}])$

Step 2.1.4

Add $0$ and $\frac{d}{d x} [- e^{- 2 x}]$ .

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

Step 2.1.5

Since $- 1$ is constant with respect to $x$ , the derivative of $- e^{- 2 x}$ with respect to $x$ is $- \frac{d}{d x} [e^{- 2 x}]$ .

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

Step 2.1.6

Multiply $- 1$ by $5$ .

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

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 $- 2 x$ .

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Differentiate.

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Step 2.3.1

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

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

Step 2.3.2

Multiply $- 2$ by $- 5$ .

$10 e^{- 2 x} \frac{d}{d x} [x]$

Step 2.3.3

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

$10 e^{- 2 x} \cdot 1$

Step 2.3.4

Multiply $10$ by $1$ .

$10 e^{- 2 x}$

Step 3

Find the second derivative of the function.

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Step 3.1

Since $10$ is constant with respect to $x$ , the derivative of $10 e^{- 2 x}$ with respect to $x$ is $10 \frac{d}{d x} [e^{- 2 x}]$ .

$f'' (x) = 10 \frac{d}{d x} (e^{- 2 x})$

Step 3.2

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 3.2.1

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

$f'' (x) = 10 (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- 2 x))$

Step 3.2.2

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

$f'' (x) = 10 (e^{u} \frac{d}{d x} (- 2 x))$

Step 3.2.3

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

$f'' (x) = 10 (e^{- 2 x} \frac{d}{d x} (- 2 x))$

Step 3.3

Differentiate.

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Step 3.3.1

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

$f'' (x) = 10 e^{- 2 x} (- 2 \frac{d}{d x} x)$

Step 3.3.2

Multiply $- 2$ by $10$ .

$f'' (x) = - 20 e^{- 2 x} \frac{d}{d x} x$

Step 3.3.3

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

$f'' (x) = - 20 e^{- 2 x} \cdot 1$

Step 3.3.4

Multiply $- 20$ by $1$ .

$f'' (x) = - 20 e^{- 2 x}$

Step 4

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$10 e^{- 2 x} = 0$

Step 5

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

No Local Extrema

Step 6

No Local Extrema

Step 7