Calculus Examples

$\frac{100}{1 + 49 e^{- 0.13 x}}$

Step 1

Differentiate using the Constant Multiple Rule.

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Step 1.1

Since $100$ is constant with respect to $x$ , the derivative of $\frac{100}{1 + 49 e^{- 0.13 x}}$ with respect to $x$ is $100 \frac{d}{d x} [\frac{1}{1 + 49 e^{- 0.13 x}}]$ .

$100 \frac{d}{d x} [\frac{1}{1 + 49 e^{- 0.13 x}}]$

Step 1.2

Rewrite $\frac{1}{1 + 49 e^{- 0.13 x}}$ as ${(1 + 49 e^{- 0.13 x})}^{- 1}$ .

$100 \frac{d}{d x} [{(1 + 49 e^{- 0.13 x})}^{- 1}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = 1 + 49 e^{- 0.13 x}$ .

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Step 2.1

To apply the Chain Rule, set $u_{1}$ as $1 + 49 e^{- 0.13 x}$ .

$100 (\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d x} [1 + 49 e^{- 0.13 x}])$

Step 2.2

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

$100 (- {u_{1}}^{- 2} \frac{d}{d x} [1 + 49 e^{- 0.13 x}])$

Step 2.3

Replace all occurrences of $u_{1}$ with $1 + 49 e^{- 0.13 x}$ .

$100 (- {(1 + 49 e^{- 0.13 x})}^{- 2} \frac{d}{d x} [1 + 49 e^{- 0.13 x}])$

Step 3

Differentiate.

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Step 3.1

Multiply $- 1$ by $100$ .

$- 100 ({(1 + 49 e^{- 0.13 x})}^{- 2} \frac{d}{d x} [1 + 49 e^{- 0.13 x}])$

Step 3.2

By the Sum Rule, the derivative of $1 + 49 e^{- 0.13 x}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [49 e^{- 0.13 x}]$ .

$- 100 {(1 + 49 e^{- 0.13 x})}^{- 2} (\frac{d}{d x} [1] + \frac{d}{d x} [49 e^{- 0.13 x}])$

Step 3.3

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

$- 100 {(1 + 49 e^{- 0.13 x})}^{- 2} (0 + \frac{d}{d x} [49 e^{- 0.13 x}])$

Step 3.4

Add $0$ and $\frac{d}{d x} [49 e^{- 0.13 x}]$ .

$- 100 {(1 + 49 e^{- 0.13 x})}^{- 2} \frac{d}{d x} [49 e^{- 0.13 x}]$

Step 3.5

Since $49$ is constant with respect to $x$ , the derivative of $49 e^{- 0.13 x}$ with respect to $x$ is $49 \frac{d}{d x} [e^{- 0.13 x}]$ .

$- 100 {(1 + 49 e^{- 0.13 x})}^{- 2} (49 \frac{d}{d x} [e^{- 0.13 x}])$

Step 3.6

Multiply $49$ by $- 100$ .

$- 4900 {(1 + 49 e^{- 0.13 x})}^{- 2} \frac{d}{d x} [e^{- 0.13 x}]$

Step 4

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 $g (x) = - 0.13 x$ .

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Step 4.1

To apply the Chain Rule, set $u_{2}$ as $- 0.13 x$ .

$- 4900 {(1 + 49 e^{- 0.13 x})}^{- 2} (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- 0.13 x])$

Step 4.2

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

$- 4900 {(1 + 49 e^{- 0.13 x})}^{- 2} (e^{u_{2}} \frac{d}{d x} [- 0.13 x])$

Step 4.3

Replace all occurrences of $u_{2}$ with $- 0.13 x$ .

$- 4900 {(1 + 49 e^{- 0.13 x})}^{- 2} (e^{- 0.13 x} \frac{d}{d x} [- 0.13 x])$

Step 5

Differentiate.

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Step 5.1

Since $- 0.13$ is constant with respect to $x$ , the derivative of $- 0.13 x$ with respect to $x$ is $- 0.13 \frac{d}{d x} [x]$ .

$- 4900 {(1 + 49 e^{- 0.13 x})}^{- 2} (e^{- 0.13 x} (- 0.13 \frac{d}{d x} [x]))$

Step 5.2

Multiply $- 0.13$ by $- 4900$ .

$637 {(1 + 49 e^{- 0.13 x})}^{- 2} (e^{- 0.13 x} (\frac{d}{d x} [x]))$

Step 5.3

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

$637 {(1 + 49 e^{- 0.13 x})}^{- 2} (e^{- 0.13 x} \cdot 1)$

Step 5.4

Multiply $e^{- 0.13 x}$ by $1$ .

$637 {(1 + 49 e^{- 0.13 x})}^{- 2} e^{- 0.13 x}$

Step 6

Simplify.

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Step 6.1

Reorder the factors of $637 {(1 + 49 e^{- 0.13 x})}^{- 2} e^{- 0.13 x}$ .

$637 e^{- 0.13 x} {(1 + 49 e^{- 0.13 x})}^{- 2}$

Step 6.2

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

$637 e^{- 0.13 x} \frac{1}{{(1 + 49 e^{- 0.13 x})}^{2}}$

Step 6.3

Multiply $637 e^{- 0.13 x} \frac{1}{{(1 + 49 e^{- 0.13 x})}^{2}}$ .

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Step 6.3.1

Combine $\frac{1}{{(1 + 49 e^{- 0.13 x})}^{2}}$ and $637$ .

$\frac{637}{{(1 + 49 e^{- 0.13 x})}^{2}} e^{- 0.13 x}$

Step 6.3.2

Combine $\frac{637}{{(1 + 49 e^{- 0.13 x})}^{2}}$ and $e^{- 0.13 x}$ .

$\frac{637 e^{- 0.13 x}}{{(1 + 49 e^{- 0.13 x})}^{2}}$