Calculus Examples

$e^{x^{2} + y} = x + y$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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 $g (x) = x^{2} + y$ .

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Step 2.1.1

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

$\frac{d}{d u} [e^{u}] \frac{d}{d x} [x^{2} + y]$

Step 2.1.2

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

$e^{u} \frac{d}{d x} [x^{2} + y]$

Step 2.1.3

Replace all occurrences of $u$ with $x^{2} + y$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

$e^{x^{2} + y} (2 x + y')$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

$\frac{d}{d x} [x] + \frac{d}{d x} [y]$

Step 3.2

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

$1 + \frac{d}{d x} [y]$

Step 3.3

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

$1 + y'$

Step 4

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

$e^{x^{2} + y} (2 x + y') = 1 + y'$

Step 5

Solve for $y'$ .

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Step 5.1

Simplify $e^{x^{2} + y} (2 x + y')$ .

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Step 5.1.1

Rewrite.

$0 + 0 + e^{x^{2} + y} (2 x + y') = 1 + y'$

Step 5.1.2

Simplify by adding zeros.

$e^{x^{2} + y} (2 x + y') = 1 + y'$

Step 5.1.3

Apply the distributive property.

$e^{x^{2} + y} (2 x) + e^{x^{2} + y} y' = 1 + y'$

Step 5.1.4

Simplify the expression.

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Step 5.1.4.1

Rewrite using the commutative property of multiplication.

$2 e^{x^{2} + y} x + e^{x^{2} + y} y' = 1 + y'$

Step 5.1.4.2

Reorder factors in $2 e^{x^{2} + y} x + e^{x^{2} + y} y'$ .

$2 x e^{x^{2} + y} + y' e^{x^{2} + y} = 1 + y'$

Step 5.2

Subtract $y'$ from both sides of the equation.

$2 x e^{x^{2} + y} + y' e^{x^{2} + y} - y' = 1$

Step 5.3

Subtract $2 x e^{x^{2} + y}$ from both sides of the equation.

$y' e^{x^{2} + y} - y' = 1 - 2 x e^{x^{2} + y}$

Step 5.4

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

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Step 5.4.1

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

$y' (e^{x^{2} + y}) - y' = 1 - 2 x e^{x^{2} + y}$

Step 5.4.2

Factor $y'$ out of $- y'$ .

$y' (e^{x^{2} + y}) + y' \cdot - 1 = 1 - 2 x e^{x^{2} + y}$

Step 5.4.3

Factor $y'$ out of $y' (e^{x^{2} + y}) + y' \cdot - 1$ .

$y' (e^{x^{2} + y} - 1) = 1 - 2 x e^{x^{2} + y}$

Step 5.5

Divide each term in $y' (e^{x^{2} + y} - 1) = 1 - 2 x e^{x^{2} + y}$ by $e^{x^{2} + y} - 1$ and simplify.

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Step 5.5.1

Divide each term in $y' (e^{x^{2} + y} - 1) = 1 - 2 x e^{x^{2} + y}$ by $e^{x^{2} + y} - 1$ .

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

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $e^{x^{2} + y} - 1$ .

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Step 5.5.2.1.1

Cancel the common factor.

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

Step 5.5.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{1}{e^{x^{2} + y} - 1} + \frac{- 2 x e^{x^{2} + y}}{e^{x^{2} + y} - 1}$

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Combine the numerators over the common denominator.

$y' = \frac{1 - 2 x e^{x^{2} + y}}{e^{x^{2} + y} - 1}$

Step 6

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

$\frac{d y}{d x} = \frac{1 - 2 x e^{x^{2} + y}}{e^{x^{2} + y} - 1}$