Calculus Examples

$y = e^{- 2 e^{- 0.01 x}}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = - 2 e^{- 0.01 x}$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $- 2 e^{- 0.01 x}$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [- 2 e^{- 0.01 x}]$

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 $a$ = $e$ .

$e^{u_{1}} \frac{d}{d x} [- 2 e^{- 0.01 x}]$

Step 1.3

Replace all occurrences of $u_{1}$ with $- 2 e^{- 0.01 x}$ .

$e^{- 2 e^{- 0.01 x}} \frac{d}{d x} [- 2 e^{- 0.01 x}]$

Step 2

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 e^{- 0.01 x}$ with respect to $x$ is $- 2 \frac{d}{d x} [e^{- 0.01 x}]$ .

$e^{- 2 e^{- 0.01 x}} (- 2 \frac{d}{d x} [e^{- 0.01 x}])$

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 $g (x) = - 0.01 x$ .

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Step 3.1

To apply the Chain Rule, set $u_{2}$ as $- 0.01 x$ .

$e^{- 2 e^{- 0.01 x}} (- 2 (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- 0.01 x]))$

Step 3.2

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

$e^{- 2 e^{- 0.01 x}} (- 2 (e^{u_{2}} \frac{d}{d x} [- 0.01 x]))$

Step 3.3

Replace all occurrences of $u_{2}$ with $- 0.01 x$ .

$e^{- 2 e^{- 0.01 x}} (- 2 (e^{- 0.01 x} \frac{d}{d x} [- 0.01 x]))$

Step 4

Multiply $e^{- 2 e^{- 0.01 x}}$ by $e^{- 0.01 x}$ by adding the exponents.

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Step 4.1

Move $e^{- 0.01 x}$ .

$e^{- 0.01 x} e^{- 2 e^{- 0.01 x}} (- 2 \frac{d}{d x} [- 0.01 x])$

Step 4.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$e^{- 0.01 x - 2 e^{- 0.01 x}} (- 2 \frac{d}{d x} [- 0.01 x])$

Step 5

Since $- 0.01$ is constant with respect to $x$ , the derivative of $- 0.01 x$ with respect to $x$ is $- 0.01 \frac{d}{d x} [x]$ .

$e^{- 0.01 x - 2 e^{- 0.01 x}} (- 2 (- 0.01 \frac{d}{d x} [x]))$

Step 6

Multiply $- 0.01$ by $- 2$ .

$e^{- 0.01 x - 2 e^{- 0.01 x}} (0.02 \frac{d}{d x} [x])$

Step 7

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

$e^{- 0.01 x - 2 e^{- 0.01 x}} (0.02 \cdot 1)$

Step 8

Simplify the expression.

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Step 8.1

Multiply $0.02$ by $1$ .

$e^{- 0.01 x - 2 e^{- 0.01 x}} \cdot 0.02$

Step 8.2

Move $0.02$ to the left of $e^{- 0.01 x - 2 e^{- 0.01 x}}$ .

$0.02 e^{- 0.01 x - 2 e^{- 0.01 x}}$