Calculus Examples

$h (x) = e^{x^{2}} \ln (x^{2})$

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 $f (x) = e^{x^{2}}$ and $g (x) = \ln (x^{2})$ .

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

Step 2

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

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

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

The derivative of $\ln (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1}}$ .

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

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

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

Step 3

Differentiate using the Power Rule.

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

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

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

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

Simplify terms.

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

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

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

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

Cancel the common factor of $x$ and $x^{2}$ .

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Factor $x$ out of $2 e^{x^{2}} x$ .

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

Cancel the common factors.

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Factor $x$ out of $x^{2}$ .

$\frac{x (2 e^{x^{2}})}{x \cdot x} + \ln (x^{2}) \frac{d}{d x} [e^{x^{2}}]$

Cancel the common factor.

$\frac{x (2 e^{x^{2}})}{x \cdot x} + \ln (x^{2}) \frac{d}{d x} [e^{x^{2}}]$

Rewrite the expression.

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

Step 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 $g (x) = x^{2}$ .

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

$\frac{2 e^{x^{2}}}{x} + \ln (x^{2}) (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [x^{2}])$

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

$\frac{2 e^{x^{2}}}{x} + \ln (x^{2}) (e^{u_{2}} \frac{d}{d x} [x^{2}])$

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

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

Step 5

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

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

Step 6

Simplify.

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Reorder terms.

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

Reorder factors in $2 e^{x^{2}} x \ln (x^{2}) + \frac{2 e^{x^{2}}}{x}$ .

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