Calculus Examples

$y = e^{- u} \cos (u)$

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d u} [f (u) g (u)]$ is $f (u) \frac{d}{d u} [g (u)] + g (u) \frac{d}{d u} [f (u)]$ where $f (u) = e^{- u}$ and $g (u) = \cos (u)$ .

$e^{- u} \frac{d}{d u} [\cos (u)] + \cos (u) \frac{d}{d u} [e^{- u}]$

Step 2

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

$e^{- u} (- \sin (u)) + \cos (u) \frac{d}{d u} [e^{- u}]$

Step 3

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

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Step 3.1

To apply the Chain Rule, set $u_{1}$ as $- u$ .

$e^{- u} (- \sin (u)) + \cos (u) (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d u} [- u])$

Step 3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{- u} (- \sin (u)) + \cos (u) (e^{u_{1}} \frac{d}{d u} [- u])$

Step 3.3

Replace all occurrences of $u_{1}$ with $- u$ .

$e^{- u} (- \sin (u)) + \cos (u) (e^{- u} \frac{d}{d u} [- u])$

Step 4

Differentiate.

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Step 4.1

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

$e^{- u} (- \sin (u)) + \cos (u) (e^{- u} (- \frac{d}{d u} [u]))$

Step 4.2

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

$e^{- u} (- \sin (u)) + \cos (u) (e^{- u} (- 1 \cdot 1))$

Step 4.3

Simplify the expression.

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Step 4.3.1

Multiply $- 1$ by $1$ .

$e^{- u} (- \sin (u)) + \cos (u) (e^{- u} \cdot - 1)$

Step 4.3.2

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

$e^{- u} (- \sin (u)) + \cos (u) (- 1 \cdot e^{- u})$

Step 4.3.3

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

$e^{- u} (- \sin (u)) + \cos (u) (- e^{- u})$

Step 4.3.4

Reorder terms.

$- e^{- u} \sin (u) - e^{- u} \cos (u)$