Calculus Examples

$y = e^{- x} \cosh (x)$

Step 1

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 $f (x) = e^{- x}$ and $g (x) = \cosh (x)$ .

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

Step 2

The derivative of $\cosh (x)$ with respect to $x$ is $\sinh (x)$ .

$e^{- x} \sinh (x) + \cosh (x) \frac{d}{d x} [e^{- x}]$

Step 3

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 $g (x) = - x$ .

Tap for more steps...

Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

$e^{- x} \sinh (x) + \cosh (x) (e^{- x} \frac{d}{d x} [- x])$

Step 4

Differentiate.

Tap for more steps...

Step 4.1

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

$e^{- x} \sinh (x) + \cosh (x) (e^{- x} (- \frac{d}{d x} [x]))$

Step 4.2

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

$e^{- x} \sinh (x) + \cosh (x) (e^{- x} (- 1 \cdot 1))$

Step 4.3

Simplify the expression.

Tap for more steps...

Step 4.3.1

Multiply $- 1$ by $1$ .

$e^{- x} \sinh (x) + \cosh (x) (e^{- x} \cdot - 1)$

Step 4.3.2

Move $- 1$ to the left of $e^{- x}$ .

$e^{- x} \sinh (x) + \cosh (x) (- 1 \cdot e^{- x})$

Step 4.3.3

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$e^{- x} \sinh (x) + \cosh (x) (- e^{- x})$

Step 4.3.4

Reorder terms.

$e^{- x} \sinh (x) - e^{- x} \cosh (x)$