Calculus Examples

$e^{- {(t - 3)}^{2}}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = e^{t}$ and $g (t) = - {(t - 3)}^{2}$ .

Tap for more steps...

Step 1.1

To apply the Chain Rule, set $u_{1}$ as $- {(t - 3)}^{2}$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d t} [- {(t - 3)}^{2}]$

Step 1.2

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

$e^{u_{1}} \frac{d}{d t} [- {(t - 3)}^{2}]$

Step 1.3

Replace all occurrences of $u_{1}$ with $- {(t - 3)}^{2}$ .

$e^{- {(t - 3)}^{2}} \frac{d}{d t} [- {(t - 3)}^{2}]$

Step 2

Since $- 1$ is constant with respect to $t$ , the derivative of $- {(t - 3)}^{2}$ with respect to $t$ is $- \frac{d}{d t} [{(t - 3)}^{2}]$ .

$e^{- {(t - 3)}^{2}} (- \frac{d}{d t} [{(t - 3)}^{2}])$

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{2}$ and $g (t) = t - 3$ .

Tap for more steps...

Step 3.1

To apply the Chain Rule, set $u_{2}$ as $t - 3$ .

$e^{- {(t - 3)}^{2}} (- (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d t} [t - 3]))$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 2$ .

$e^{- {(t - 3)}^{2}} (- (2 u_{2} \frac{d}{d t} [t - 3]))$

Step 3.3

Replace all occurrences of $u_{2}$ with $t - 3$ .

$e^{- {(t - 3)}^{2}} (- (2 (t - 3) \frac{d}{d t} [t - 3]))$

Step 4

Differentiate.

Tap for more steps...

Step 4.1

Multiply $2$ by $- 1$ .

$e^{- {(t - 3)}^{2}} (- 2 ((t - 3) \frac{d}{d t} [t - 3]))$

Step 4.2

By the Sum Rule, the derivative of $t - 3$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [- 3]$ .

$e^{- {(t - 3)}^{2}} (- 2 (t - 3) (\frac{d}{d t} [t] + \frac{d}{d t} [- 3]))$

Step 4.3

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

$e^{- {(t - 3)}^{2}} (- 2 (t - 3) (1 + \frac{d}{d t} [- 3]))$

Step 4.4

Since $- 3$ is constant with respect to $t$ , the derivative of $- 3$ with respect to $t$ is $0$ .

$e^{- {(t - 3)}^{2}} (- 2 (t - 3) (1 + 0))$

Step 4.5

Simplify the expression.

Tap for more steps...

Step 4.5.1

Add $1$ and $0$ .

$e^{- {(t - 3)}^{2}} (- 2 (t - 3) \cdot 1)$

Step 4.5.2

Multiply $- 2$ by $1$ .

$e^{- {(t - 3)}^{2}} (- 2 (t - 3))$

Step 5

Simplify.

Tap for more steps...

Step 5.1

Apply the distributive property.

$e^{- {(t - 3)}^{2}} (- 2 t - 2 \cdot - 3)$

Step 5.2

Multiply $- 2$ by $- 3$ .

$e^{- {(t - 3)}^{2}} (- 2 t + 6)$