Calculus Examples

$x e^{y} - 10 x + 3 y = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x e^{y} - 10 x + 3 y) = \frac{d}{d x} (0)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $x e^{y} - 10 x + 3 y$ with respect to $x$ is $\frac{d}{d x} [x e^{y}] + \frac{d}{d x} [- 10 x] + \frac{d}{d x} [3 y]$ .

$\frac{d}{d x} [x e^{y}] + \frac{d}{d x} [- 10 x] + \frac{d}{d x} [3 y]$

Step 2.2

Evaluate $\frac{d}{d x} [x e^{y}]$ .

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Step 2.2.1

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 $f (x) = x$ and $g (x) = e^{y}$ .

$x \frac{d}{d x} [e^{y}] + e^{y} \frac{d}{d x} [x] + \frac{d}{d x} [- 10 x] + \frac{d}{d x} [3 y]$

Step 2.2.2

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

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Step 2.2.2.1

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

$x (\frac{d}{d u} [e^{u}] \frac{d}{d x} [y]) + e^{y} \frac{d}{d x} [x] + \frac{d}{d x} [- 10 x] + \frac{d}{d x} [3 y]$

Step 2.2.2.2

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

$x (e^{u} \frac{d}{d x} [y]) + e^{y} \frac{d}{d x} [x] + \frac{d}{d x} [- 10 x] + \frac{d}{d x} [3 y]$

Step 2.2.2.3

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

$x (e^{y} \frac{d}{d x} [y]) + e^{y} \frac{d}{d x} [x] + \frac{d}{d x} [- 10 x] + \frac{d}{d x} [3 y]$

Step 2.2.3

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$x (e^{y} y') + e^{y} \frac{d}{d x} [x] + \frac{d}{d x} [- 10 x] + \frac{d}{d x} [3 y]$

Step 2.2.4

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

$x (e^{y} y') + e^{y} \cdot 1 + \frac{d}{d x} [- 10 x] + \frac{d}{d x} [3 y]$

Step 2.2.5

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

$x e^{y} y' + e^{y} + \frac{d}{d x} [- 10 x] + \frac{d}{d x} [3 y]$

Step 2.3

Evaluate $\frac{d}{d x} [- 10 x]$ .

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Step 2.3.1

Since $- 10$ is constant with respect to $x$ , the derivative of $- 10 x$ with respect to $x$ is $- 10 \frac{d}{d x} [x]$ .

$x e^{y} y' + e^{y} - 10 \frac{d}{d x} [x] + \frac{d}{d x} [3 y]$

Step 2.3.2

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

$x e^{y} y' + e^{y} - 10 \cdot 1 + \frac{d}{d x} [3 y]$

Step 2.3.3

Multiply $- 10$ by $1$ .

$x e^{y} y' + e^{y} - 10 + \frac{d}{d x} [3 y]$

Step 2.4

Evaluate $\frac{d}{d x} [3 y]$ .

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Step 2.4.1

Since $3$ is constant with respect to $x$ , the derivative of $3 y$ with respect to $x$ is $3 \frac{d}{d x} [y]$ .

$x e^{y} y' + e^{y} - 10 + 3 \frac{d}{d x} [y]$

Step 2.4.2

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$x e^{y} y' + e^{y} - 10 + 3 y'$

Step 2.5

Simplify.

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Step 2.5.1

Reorder terms.

$e^{y} x y' + 3 y' + e^{y} - 10$

Step 2.5.2

Reorder factors in $e^{y} x y' + 3 y' + e^{y} - 10$ .

$x e^{y} y' + 3 y' + e^{y} - 10$

Step 3

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

$0$

Step 4

Reform the equation by setting the left side equal to the right side.

$x e^{y} y' + 3 y' + e^{y} - 10 = 0$

Step 5

Solve for $y'$ .

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Step 5.1

Reorder factors in $x e^{y} y' + 3 y' + e^{y} - 10$ .

$x y' e^{y} + 3 y' + e^{y} - 10 = 0$

Step 5.2

Move all terms not containing $y'$ to the right side of the equation.

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Step 5.2.1

Subtract $e^{y}$ from both sides of the equation.

$x y' e^{y} + 3 y' - 10 = - e^{y}$

Step 5.2.2

Add $10$ to both sides of the equation.

$x y' e^{y} + 3 y' = - e^{y} + 10$

Step 5.3

Factor $y'$ out of $x y' e^{y} + 3 y'$ .

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Step 5.3.1

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

$y' (x e^{y}) + 3 y' = - e^{y} + 10$

Step 5.3.2

Factor $y'$ out of $3 y'$ .

$y' (x e^{y}) + y' \cdot 3 = - e^{y} + 10$

Step 5.3.3

Factor $y'$ out of $y' (x e^{y}) + y' \cdot 3$ .

$y' (x e^{y} + 3) = - e^{y} + 10$

Step 5.4

Divide each term in $y' (x e^{y} + 3) = - e^{y} + 10$ by $x e^{y} + 3$ and simplify.

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Step 5.4.1

Divide each term in $y' (x e^{y} + 3) = - e^{y} + 10$ by $x e^{y} + 3$ .

$\frac{y' (x e^{y} + 3)}{x e^{y} + 3} = \frac{- e^{y}}{x e^{y} + 3} + \frac{10}{x e^{y} + 3}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $x e^{y} + 3$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{y' (x e^{y} + 3)}{x e^{y} + 3} = \frac{- e^{y}}{x e^{y} + 3} + \frac{10}{x e^{y} + 3}$

Step 5.4.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- e^{y}}{x e^{y} + 3} + \frac{10}{x e^{y} + 3}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Combine the numerators over the common denominator.

$y' = \frac{- e^{y} + 10}{x e^{y} + 3}$

Step 5.4.3.2

Factor $- 1$ out of $- e^{y}$ .

$y' = \frac{- (e^{y}) + 10}{x e^{y} + 3}$

Step 5.4.3.3

Rewrite $10$ as $- 1 (- 10)$ .

$y' = \frac{- (e^{y}) - 1 (- 10)}{x e^{y} + 3}$

Step 5.4.3.4

Factor $- 1$ out of $- (e^{y}) - 1 (- 10)$ .

$y' = \frac{- (e^{y} - 10)}{x e^{y} + 3}$

Step 5.4.3.5

Simplify the expression.

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Step 5.4.3.5.1

Rewrite $- (e^{y} - 10)$ as $- 1 (e^{y} - 10)$ .

$y' = \frac{- 1 (e^{y} - 10)}{x e^{y} + 3}$

Step 5.4.3.5.2

Move the negative in front of the fraction.

$y' = - \frac{e^{y} - 10}{x e^{y} + 3}$

Step 6

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = - \frac{e^{y} - 10}{x e^{y} + 3}$