Calculus Examples

$\sin (h (x)) = \frac{e^{x} - e^{- x}}{2}$

Step 1

Multiply $h$ by $x$ .

$\sin (h x) = \frac{e^{x} - e^{- x}}{2}$

Step 2

Differentiate both sides of the equation.

$\frac{d}{d x} (\sin (h x)) = \frac{d}{d x} (\frac{e^{x} - e^{- x}}{2})$

Step 3

Differentiate the left side of the equation.

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Step 3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \sin (x)$ and $g (x) = h x$ .

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Step 3.1.1

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

$\frac{d}{d u} [\sin (u)] \frac{d}{d x} [h x]$

Step 3.1.2

The derivative of $\sin (u)$ with respect to $u$ is $\cos (u)$ .

$\cos (u) \frac{d}{d x} [h x]$

Step 3.1.3

Replace all occurrences of $u$ with $h x$ .

$\cos (h x) \frac{d}{d x} [h x]$

Step 3.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = h$ and $g (x) = x$ .

$\cos (h x) (h \frac{d}{d x} [x] + x \frac{d}{d x} [h])$

Step 3.3

Differentiate using the Power Rule.

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Step 3.3.1

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

$\cos (h x) (h \cdot 1 + x \frac{d}{d x} [h])$

Step 3.3.2

Multiply $h$ by $1$ .

$\cos (h x) (h + x \frac{d}{d x} [h])$

Step 3.4

Rewrite $\frac{d}{d x} [h]$ as $h'$ .

$\cos (h x) (h + x h')$

Step 3.5

Simplify.

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Step 3.5.1

Apply the distributive property.

$\cos (h x) h + \cos (h x) (x h')$

Step 3.5.2

Reorder terms.

$h \cos (h x) + x \cos (h x) h'$

Step 4

Differentiate the right side of the equation.

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Step 4.1

Differentiate.

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Step 4.1.1

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

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

Step 4.1.2

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

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

Step 4.2

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

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

Step 4.3

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

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

Step 4.4

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 $g (x) = - x$ .

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Step 4.4.1

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

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

Step 4.4.2

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

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

Step 4.4.3

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

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

Step 4.5

Differentiate.

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Step 4.5.1

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

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

Step 4.5.2

Multiply.

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Step 4.5.2.1

Multiply $- 1$ by $- 1$ .

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

Step 4.5.2.2

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

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

Step 4.5.3

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

$\frac{1}{2} (e^{x} + e^{- x} \cdot 1)$

Step 4.5.4

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

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

Step 4.6

Simplify.

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Step 4.6.1

Apply the distributive property.

$\frac{1}{2} e^{x} + \frac{1}{2} e^{- x}$

Step 4.6.2

Simplify each term.

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Step 4.6.2.1

Combine $\frac{1}{2}$ and $e^{x}$ .

$\frac{e^{x}}{2} + \frac{1}{2} e^{- x}$

Step 4.6.2.2

Combine $\frac{1}{2}$ and $e^{- x}$ .

$\frac{e^{x}}{2} + \frac{e^{- x}}{2}$

Step 5

Reform the equation by setting the left side equal to the right side.

$h \cos (h x) + x \cos (h x) h' = \frac{e^{x}}{2} + \frac{e^{- x}}{2}$

Step 6

Solve for $h'$ .

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Step 6.1

Simplify the left side.

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Step 6.1.1

Reorder factors in $h \cos (h x) + x \cos (h x) h'$ .

$h \cos (h x) + x h' \cos (h x) = \frac{e^{x}}{2} + \frac{e^{- x}}{2}$

Step 6.2

Subtract $h \cos (h x)$ from both sides of the equation.

$x h' \cos (h x) = \frac{e^{x}}{2} + \frac{e^{- x}}{2} - h \cos (h x)$

Step 6.3

Divide each term in $x h' \cos (h x) = \frac{e^{x}}{2} + \frac{e^{- x}}{2} - h \cos (h x)$ by $x \cos (h x)$ and simplify.

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Step 6.3.1

Divide each term in $x h' \cos (h x) = \frac{e^{x}}{2} + \frac{e^{- x}}{2} - h \cos (h x)$ by $x \cos (h x)$ .

$\frac{x h' \cos (h x)}{x \cos (h x)} = \frac{\frac{e^{x}}{2}}{x \cos (h x)} + \frac{\frac{e^{- x}}{2}}{x \cos (h x)} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.2

Simplify the left side.

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Step 6.3.2.1

Cancel the common factor of $x$ .

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Step 6.3.2.1.1

Cancel the common factor.

$\frac{x h' \cos (h x)}{x \cos (h x)} = \frac{\frac{e^{x}}{2}}{x \cos (h x)} + \frac{\frac{e^{- x}}{2}}{x \cos (h x)} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.2.1.2

Rewrite the expression.

$\frac{h' \cos (h x)}{\cos (h x)} = \frac{\frac{e^{x}}{2}}{x \cos (h x)} + \frac{\frac{e^{- x}}{2}}{x \cos (h x)} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.2.2

Cancel the common factor of $\cos (h x)$ .

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Step 6.3.2.2.1

Cancel the common factor.

$\frac{h' \cos (h x)}{\cos (h x)} = \frac{\frac{e^{x}}{2}}{x \cos (h x)} + \frac{\frac{e^{- x}}{2}}{x \cos (h x)} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.2.2.2

Divide $h'$ by $1$ .

$h' = \frac{\frac{e^{x}}{2}}{x \cos (h x)} + \frac{\frac{e^{- x}}{2}}{x \cos (h x)} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3

Simplify the right side.

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Step 6.3.3.1

Simplify each term.

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Step 6.3.3.1.1

Separate fractions.

$h' = \frac{\frac{e^{x}}{2}}{x} \cdot \frac{1}{\cos (h x)} + \frac{\frac{e^{- x}}{2}}{x \cos (h x)} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.2

Convert from $\frac{1}{\cos (h x)}$ to $\sec (h x)$ .

$h' = \frac{\frac{e^{x}}{2}}{x} \sec (h x) + \frac{\frac{e^{- x}}{2}}{x \cos (h x)} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.3

Multiply the numerator by the reciprocal of the denominator.

$h' = \frac{e^{x}}{2} \cdot \frac{1}{x} \sec (h x) + \frac{\frac{e^{- x}}{2}}{x \cos (h x)} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.4

Combine.

$h' = \frac{e^{x} \cdot 1}{2 x} \sec (h x) + \frac{\frac{e^{- x}}{2}}{x \cos (h x)} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.5

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

$h' = \frac{e^{x}}{2 x} \sec (h x) + \frac{\frac{e^{- x}}{2}}{x \cos (h x)} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.6

Combine $\frac{e^{x}}{2 x}$ and $\sec (h x)$ .

$h' = \frac{e^{x} \sec (h x)}{2 x} + \frac{\frac{e^{- x}}{2}}{x \cos (h x)} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.7

Separate fractions.

$h' = \frac{e^{x} \sec (h x)}{2 x} + \frac{\frac{e^{- x}}{2}}{x} \cdot \frac{1}{\cos (h x)} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.8

Convert from $\frac{1}{\cos (h x)}$ to $\sec (h x)$ .

$h' = \frac{e^{x} \sec (h x)}{2 x} + \frac{\frac{e^{- x}}{2}}{x} \sec (h x) + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.9

Multiply the numerator by the reciprocal of the denominator.

$h' = \frac{e^{x} \sec (h x)}{2 x} + \frac{e^{- x}}{2} \cdot \frac{1}{x} \sec (h x) + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.10

Combine.

$h' = \frac{e^{x} \sec (h x)}{2 x} + \frac{e^{- x} \cdot 1}{2 x} \sec (h x) + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.11

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

$h' = \frac{e^{x} \sec (h x)}{2 x} + \frac{e^{- x}}{2 x} \sec (h x) + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.12

Combine $\frac{e^{- x}}{2 x}$ and $\sec (h x)$ .

$h' = \frac{e^{x} \sec (h x)}{2 x} + \frac{e^{- x} \sec (h x)}{2 x} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.13

Cancel the common factor of $\cos (h x)$ .

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Step 6.3.3.1.13.1

Cancel the common factor.

$h' = \frac{e^{x} \sec (h x)}{2 x} + \frac{e^{- x} \sec (h x)}{2 x} + \frac{- h \cos (h x)}{x \cos (h x)}$

Step 6.3.3.1.13.2

Rewrite the expression.

$h' = \frac{e^{x} \sec (h x)}{2 x} + \frac{e^{- x} \sec (h x)}{2 x} + \frac{- h}{x}$

Step 6.3.3.1.14

Move the negative in front of the fraction.

$h' = \frac{e^{x} \sec (h x)}{2 x} + \frac{e^{- x} \sec (h x)}{2 x} - \frac{h}{x}$

Step 7

Replace $h'$ with $\frac{d h}{d x}$ .

$\frac{d h}{d x} = \frac{e^{x} \sec (h x)}{2 x} + \frac{e^{- x} \sec (h x)}{2 x} - \frac{h}{x}$