Calculus Examples

$f (x) = 3 e^{1 - x^{2}} \cdot \ln (x)$

Step 1

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

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

Step 2

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 $f (x) = e^{1 - x^{2}}$ and $g (x) = \ln (x)$ .

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

Step 3

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

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

Step 4

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

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

Step 5

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 $g (x) = 1 - x^{2}$ .

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

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

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

Step 6

Differentiate.

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Step 6.1

By the Sum Rule, the derivative of $1 - x^{2}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- x^{2}]$ .

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

Step 6.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$3 (\frac{e^{1 - x^{2}}}{x} + \ln (x) (e^{1 - x^{2}} (0 + \frac{d}{d x} [- x^{2}])))$

Step 6.3

Add $0$ and $\frac{d}{d x} [- x^{2}]$ .

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

Step 6.4

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

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

Step 6.5

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

$3 (\frac{e^{1 - x^{2}}}{x} + \ln (x) (e^{1 - x^{2}} (- (2 x))))$

Step 6.6

Multiply $2$ by $- 1$ .

$3 (\frac{e^{1 - x^{2}}}{x} + \ln (x) (e^{1 - x^{2}} (- 2 x)))$

Step 7

To write $\ln (x) (e^{1 - x^{2}} (- 2 x))$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$3 (\frac{e^{1 - x^{2}}}{x} + \frac{\ln (x) (e^{1 - x^{2}} (- 2 x)) x}{x})$

Step 8

Combine the numerators over the common denominator.

$3 \frac{e^{1 - x^{2}} + \ln (x) (e^{1 - x^{2}} (- 2 x)) x}{x}$

Step 9

Raise $x$ to the power of $1$ .

$3 \frac{e^{1 - x^{2}} + \ln (x) (e^{1 - x^{2}} (- 2 (x^{1} x)))}{x}$

Step 10

Raise $x$ to the power of $1$ .

$3 \frac{e^{1 - x^{2}} + \ln (x) (e^{1 - x^{2}} (- 2 (x^{1} x^{1})))}{x}$

Step 11

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

$3 \frac{e^{1 - x^{2}} + \ln (x) (e^{1 - x^{2}} (- 2 x^{1 + 1}))}{x}$

Step 12

Add $1$ and $1$ .

$3 \frac{e^{1 - x^{2}} + \ln (x) (e^{1 - x^{2}} (- 2 x^{2}))}{x}$

Step 13

Combine $3$ and $\frac{e^{1 - x^{2}} + \ln (x) (e^{1 - x^{2}} (- 2 x^{2}))}{x}$ .

$\frac{3 (e^{1 - x^{2}} + \ln (x) (e^{1 - x^{2}} (- 2 x^{2})))}{x}$

Step 14

Simplify.

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Step 14.1

Apply the distributive property.

$\frac{3 e^{1 - x^{2}} + 3 (\ln (x) (e^{1 - x^{2}} (- 2 x^{2})))}{x}$

Step 14.2

Simplify the numerator.

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Step 14.2.1

Simplify each term.

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Step 14.2.1.1

Rewrite using the commutative property of multiplication.

$\frac{3 e^{1 - x^{2}} + 3 (- 2 \ln (x) (e^{1 - x^{2}} x^{2}))}{x}$

Step 14.2.1.2

Simplify $- 2 \ln (x)$ by moving $2$ inside the logarithm.

$\frac{3 e^{1 - x^{2}} + 3 (- \ln (x^{2}) (e^{1 - x^{2}} x^{2}))}{x}$

Step 14.2.1.3

Multiply $3 (- \ln (x^{2}) (e^{1 - x^{2}} x^{2}))$ .

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Step 14.2.1.3.1

Multiply $- 1$ by $3$ .

$\frac{3 e^{1 - x^{2}} - 3 (\ln (x^{2}) (e^{1 - x^{2}} x^{2}))}{x}$

Step 14.2.1.3.2

Reorder $\ln (x^{2})$ and $- 3$ .

$\frac{3 e^{1 - x^{2}} - 3 \cdot \ln (x^{2}) (e^{1 - x^{2}} x^{2})}{x}$

Step 14.2.1.3.3

Simplify $- 3 \cdot \ln (x^{2})$ by moving $3$ inside the logarithm.

$\frac{3 e^{1 - x^{2}} - \ln ({(x^{2})}^{3}) (e^{1 - x^{2}} x^{2})}{x}$

Step 14.2.1.4

Multiply the exponents in ${(x^{2})}^{3}$ .

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Step 14.2.1.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{3 e^{1 - x^{2}} - \ln (x^{2 \cdot 3}) (e^{1 - x^{2}} x^{2})}{x}$

Step 14.2.1.4.2

Multiply $2$ by $3$ .

$\frac{3 e^{1 - x^{2}} - \ln (x^{6}) (e^{1 - x^{2}} x^{2})}{x}$

Step 14.2.2

Reorder factors in $3 e^{1 - x^{2}} - \ln (x^{6}) (e^{1 - x^{2}} x^{2})$ .

$\frac{3 e^{1 - x^{2}} - x^{2} e^{1 - x^{2}} \ln (x^{6})}{x}$

Step 14.3

Reorder terms.

$\frac{- e^{1 - x^{2}} x^{2} \ln (x^{6}) + 3 e^{1 - x^{2}}}{x}$

Step 14.4

Factor $e^{1 - x^{2}}$ out of $- e^{1 - x^{2}} x^{2} \ln (x^{6}) + 3 e^{1 - x^{2}}$ .

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Step 14.4.1

Factor $e^{1 - x^{2}}$ out of $- e^{1 - x^{2}} x^{2} \ln (x^{6})$ .

$\frac{e^{1 - x^{2}} (- x^{2} \ln (x^{6})) + 3 e^{1 - x^{2}}}{x}$

Step 14.4.2

Factor $e^{1 - x^{2}}$ out of $3 e^{1 - x^{2}}$ .

$\frac{e^{1 - x^{2}} (- x^{2} \ln (x^{6})) + e^{1 - x^{2}} \cdot 3}{x}$

Step 14.4.3

Factor $e^{1 - x^{2}}$ out of $e^{1 - x^{2}} (- x^{2} \ln (x^{6})) + e^{1 - x^{2}} \cdot 3$ .

$\frac{e^{1 - x^{2}} (- x^{2} \ln (x^{6}) + 3)}{x}$

Step 14.5

Factor $- 1$ out of $- x^{2} \ln (x^{6})$ .

$\frac{e^{1 - x^{2}} (- (x^{2} \ln (x^{6})) + 3)}{x}$

Step 14.6

Rewrite $3$ as $- 1 (- 3)$ .

$\frac{e^{1 - x^{2}} (- (x^{2} \ln (x^{6})) - 1 (- 3))}{x}$

Step 14.7

Factor $- 1$ out of $- (x^{2} \ln (x^{6})) - 1 (- 3)$ .

$\frac{e^{1 - x^{2}} (- (x^{2} \ln (x^{6}) - 3))}{x}$

Step 14.8

Rewrite $- (x^{2} \ln (x^{6}) - 3)$ as $- 1 (x^{2} \ln (x^{6}) - 3)$ .

$\frac{e^{1 - x^{2}} (- 1 (x^{2} \ln (x^{6}) - 3))}{x}$

Step 14.9

Move the negative in front of the fraction.

$- \frac{e^{1 - x^{2}} (x^{2} \ln (x^{6}) - 3)}{x}$