Calculus Examples

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

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d y} [\frac{f (y)}{g (y)}]$ is $\frac{g (y) \frac{d}{d y} [f (y)] - f (y) \frac{d}{d y} [g (y)]}{{g (y)}^{2}}$ where $f (y) = y e^{x} + x e^{x} - e^{x}$ and $g (y) = {(y + x)}^{2}$ .

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

Step 2

Differentiate.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Multiply $2$ by $2$ .

$\frac{{(y + x)}^{2} \frac{d}{d y} [y e^{x} + x e^{x} - e^{x}] - (y e^{x} + x e^{x} - e^{x}) \frac{d}{d y} [{(y + x)}^{2}]}{{(y + x)}^{4}}$

Step 2.2

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

$\frac{{(y + x)}^{2} (\frac{d}{d y} [y e^{x}] + \frac{d}{d y} [x e^{x}] + \frac{d}{d y} [- e^{x}]) - (y e^{x} + x e^{x} - e^{x}) \frac{d}{d y} [{(y + x)}^{2}]}{{(y + x)}^{4}}$

Step 2.3

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

$\frac{{(y + x)}^{2} (e^{x} \frac{d}{d y} [y] + \frac{d}{d y} [x e^{x}] + \frac{d}{d y} [- e^{x}]) - (y e^{x} + x e^{x} - e^{x}) \frac{d}{d y} [{(y + x)}^{2}]}{{(y + x)}^{4}}$

Step 2.4

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

$\frac{{(y + x)}^{2} (e^{x} \cdot 1 + \frac{d}{d y} [x e^{x}] + \frac{d}{d y} [- e^{x}]) - (y e^{x} + x e^{x} - e^{x}) \frac{d}{d y} [{(y + x)}^{2}]}{{(y + x)}^{4}}$

Step 2.5

Multiply $e^{x}$ by $1$ .

$\frac{{(y + x)}^{2} (e^{x} + \frac{d}{d y} [x e^{x}] + \frac{d}{d y} [- e^{x}]) - (y e^{x} + x e^{x} - e^{x}) \frac{d}{d y} [{(y + x)}^{2}]}{{(y + x)}^{4}}$

Step 2.6

Since $x e^{x}$ is constant with respect to $y$ , the derivative of $x e^{x}$ with respect to $y$ is $0$ .

$\frac{{(y + x)}^{2} (e^{x} + 0 + \frac{d}{d y} [- e^{x}]) - (y e^{x} + x e^{x} - e^{x}) \frac{d}{d y} [{(y + x)}^{2}]}{{(y + x)}^{4}}$

Step 2.7

Add $e^{x}$ and $0$ .

$\frac{{(y + x)}^{2} (e^{x} + \frac{d}{d y} [- e^{x}]) - (y e^{x} + x e^{x} - e^{x}) \frac{d}{d y} [{(y + x)}^{2}]}{{(y + x)}^{4}}$

Step 2.8

Since $- e^{x}$ is constant with respect to $y$ , the derivative of $- e^{x}$ with respect to $y$ is $0$ .

$\frac{{(y + x)}^{2} (e^{x} + 0) - (y e^{x} + x e^{x} - e^{x}) \frac{d}{d y} [{(y + x)}^{2}]}{{(y + x)}^{4}}$

Step 2.9

Add $e^{x}$ and $0$ .

$\frac{{(y + x)}^{2} e^{x} - (y e^{x} + x e^{x} - e^{x}) \frac{d}{d y} [{(y + x)}^{2}]}{{(y + x)}^{4}}$

Step 3

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

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Step 3.1

To apply the Chain Rule, set $u$ as $y + x$ .

$\frac{{(y + x)}^{2} e^{x} - (y e^{x} + x e^{x} - e^{x}) (\frac{d}{d u} [u^{2}] \frac{d}{d y} [y + x])}{{(y + x)}^{4}}$

Step 3.2

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

$\frac{{(y + x)}^{2} e^{x} - (y e^{x} + x e^{x} - e^{x}) (2 u \frac{d}{d y} [y + x])}{{(y + x)}^{4}}$

Step 3.3

Replace all occurrences of $u$ with $y + x$ .

$\frac{{(y + x)}^{2} e^{x} - (y e^{x} + x e^{x} - e^{x}) (2 (y + x) \frac{d}{d y} [y + x])}{{(y + x)}^{4}}$

Step 4

Simplify with factoring out.

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Step 4.1

Multiply $2$ by $- 1$ .

$\frac{{(y + x)}^{2} e^{x} - 2 (y e^{x} + x e^{x} - e^{x}) ((y + x) \frac{d}{d y} [y + x])}{{(y + x)}^{4}}$

Step 4.2

Factor $y + x$ out of ${(y + x)}^{2} e^{x} - 2 (y e^{x} + x e^{x} - e^{x}) ((y + x) \frac{d}{d y} [y + x])$ .

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Step 4.2.1

Factor $y + x$ out of ${(y + x)}^{2} e^{x}$ .

$\frac{(y + x) ((y + x) e^{x}) - 2 (y e^{x} + x e^{x} - e^{x}) ((y + x) \frac{d}{d y} [y + x])}{{(y + x)}^{4}}$

Step 4.2.2

Factor $y + x$ out of $- 2 (y e^{x} + x e^{x} - e^{x}) ((y + x) \frac{d}{d y} [y + x])$ .

$\frac{(y + x) ((y + x) e^{x}) + (y + x) (- 2 (y e^{x} + x e^{x} - e^{x}) (\frac{d}{d y} [y + x]))}{{(y + x)}^{4}}$

Step 4.2.3

Factor $y + x$ out of $(y + x) ((y + x) e^{x}) + (y + x) (- 2 (y e^{x} + x e^{x} - e^{x}) (\frac{d}{d y} [y + x]))$ .

$\frac{(y + x) ((y + x) e^{x} - 2 (y e^{x} + x e^{x} - e^{x}) (\frac{d}{d y} [y + x]))}{{(y + x)}^{4}}$

Step 5

Cancel the common factors.

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Step 5.1

Factor $y + x$ out of ${(y + x)}^{4}$ .

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

Step 5.2

Cancel the common factor.

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

Step 5.3

Rewrite the expression.

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

Step 6

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

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

Step 7

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

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

Step 8

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

$\frac{(y + x) e^{x} - 2 (y e^{x} + x e^{x} - e^{x}) (1 + 0)}{{(y + x)}^{3}}$

Step 9

Simplify the expression.

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Step 9.1

Add $1$ and $0$ .

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

Step 9.2

Multiply $- 2$ by $1$ .

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

Step 10

Simplify.

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Step 10.1

Apply the distributive property.

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

Step 10.2

Apply the distributive property.

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

Step 10.3

Simplify the numerator.

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Step 10.3.1

Multiply $- 1$ by $- 2$ .

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

Step 10.3.2

Subtract $2 y e^{x}$ from $y e^{x}$ .

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

Step 10.3.3

Subtract $2 x e^{x}$ from $x e^{x}$ .

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

Step 10.4

Reorder terms.

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

Step 10.5

Factor $e^{x}$ out of $- e^{x} y - e^{x} x + 2 e^{x}$ .

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Step 10.5.1

Factor $e^{x}$ out of $- e^{x} y$ .

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

Step 10.5.2

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

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

Step 10.5.3

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

$\frac{e^{x} (- y) + e^{x} (- x) + e^{x} \cdot 2}{{(x + y)}^{3}}$

Step 10.5.4

Factor $e^{x}$ out of $e^{x} (- y) + e^{x} (- x)$ .

$\frac{e^{x} (- y - x) + e^{x} \cdot 2}{{(x + y)}^{3}}$

Step 10.5.5

Factor $e^{x}$ out of $e^{x} (- y - x) + e^{x} \cdot 2$ .

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

Step 10.6

Factor $- 1$ out of $- y$ .

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

Step 10.7

Factor $- 1$ out of $- x$ .

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

Step 10.8

Factor $- 1$ out of $- (y) - (x)$ .

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

Step 10.9

Rewrite $2$ as $- 1 (- 2)$ .

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

Step 10.10

Factor $- 1$ out of $- (y + x) - 1 (- 2)$ .

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

Step 10.11

Rewrite $- (y + x - 2)$ as $- 1 (y + x - 2)$ .

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

Step 10.12

Move the negative in front of the fraction.

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