Calculus Examples

$e^{x y} = e^{4 x} - e^{5 y}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (e^{x y}) = \frac{d}{d y} (e^{4 x} - e^{5 y})$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

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

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

Step 2.2

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

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

Step 2.3

Differentiate using the Power Rule.

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Step 2.3.1

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

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

Step 2.3.2

Multiply $x$ by $1$ .

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

Step 2.4

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

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

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Reorder terms.

$e^{x y} x + e^{x y} y x'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $e^{4 x} - e^{5 y}$ with respect to $y$ is $\frac{d}{d y} [e^{4 x}] + \frac{d}{d y} [- e^{5 y}]$ .

$\frac{d}{d y} [e^{4 x}] + \frac{d}{d y} [- e^{5 y}]$

Step 3.2

Evaluate $\frac{d}{d y} [e^{4 x}]$ .

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Step 3.2.1

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = e^{y}$ and $g (y) = 4 x$ .

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Step 3.2.1.1

To apply the Chain Rule, set $u_{1}$ as $4 x$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d y} [4 x] + \frac{d}{d y} [- e^{5 y}]$

Step 3.2.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{u_{1}} \frac{d}{d y} [4 x] + \frac{d}{d y} [- e^{5 y}]$

Step 3.2.1.3

Replace all occurrences of $u_{1}$ with $4 x$ .

$e^{4 x} \frac{d}{d y} [4 x] + \frac{d}{d y} [- e^{5 y}]$

Step 3.2.2

Since $4$ is constant with respect to $y$ , the derivative of $4 x$ with respect to $y$ is $4 \frac{d}{d y} [x]$ .

$e^{4 x} (4 \frac{d}{d y} [x]) + \frac{d}{d y} [- e^{5 y}]$

Step 3.2.3

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

$e^{4 x} (4 x') + \frac{d}{d y} [- e^{5 y}]$

Step 3.3

Evaluate $\frac{d}{d y} [- e^{5 y}]$ .

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Step 3.3.1

Since $- 1$ is constant with respect to $y$ , the derivative of $- e^{5 y}$ with respect to $y$ is $- \frac{d}{d y} [e^{5 y}]$ .

$e^{4 x} (4 x') - \frac{d}{d y} [e^{5 y}]$

Step 3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = e^{y}$ and $g (y) = 5 y$ .

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Step 3.3.2.1

To apply the Chain Rule, set $u_{2}$ as $5 y$ .

$e^{4 x} (4 x') - (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d y} [5 y])$

Step 3.3.2.2

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

$e^{4 x} (4 x') - (e^{u_{2}} \frac{d}{d y} [5 y])$

Step 3.3.2.3

Replace all occurrences of $u_{2}$ with $5 y$ .

$e^{4 x} (4 x') - (e^{5 y} \frac{d}{d y} [5 y])$

Step 3.3.3

Since $5$ is constant with respect to $y$ , the derivative of $5 y$ with respect to $y$ is $5 \frac{d}{d y} [y]$ .

$e^{4 x} (4 x') - (e^{5 y} (5 \frac{d}{d y} [y]))$

Step 3.3.4

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

$e^{4 x} (4 x') - (e^{5 y} (5 \cdot 1))$

Step 3.3.5

Multiply $5$ by $1$ .

$e^{4 x} (4 x') - (e^{5 y} \cdot 5)$

Step 3.3.6

Move $5$ to the left of $e^{5 y}$ .

$e^{4 x} (4 x') - (5 \cdot e^{5 y})$

Step 3.3.7

Multiply $5$ by $- 1$ .

$e^{4 x} (4 x') - 5 e^{5 y}$

Step 3.4

Reorder terms.

$4 e^{4 x} x' - 5 e^{5 y}$

Step 4

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

$e^{x y} x + e^{x y} y x' = 4 e^{4 x} x' - 5 e^{5 y}$

Step 5

Solve for $x'$ .

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Step 5.1

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

$x e^{x y} + y x' e^{x y} = 4 e^{4 x} x' - 5 e^{5 y}$

Step 5.2

Reorder factors in $4 e^{4 x} x' - 5 e^{5 y}$ .

$x e^{x y} + y x' e^{x y} = 4 x' e^{4 x} - 5 e^{5 y}$

Step 5.3

Subtract $4 x' e^{4 x}$ from both sides of the equation.

$x e^{x y} + y x' e^{x y} - 4 x' e^{4 x} = - 5 e^{5 y}$

Step 5.4

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

$y x' e^{x y} - 4 x' e^{4 x} = - 5 e^{5 y} - x e^{x y}$

Step 5.5

Factor $x'$ out of $y x' e^{x y} - 4 x' e^{4 x}$ .

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Step 5.5.1

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

$x' (y e^{x y}) - 4 x' e^{4 x} = - 5 e^{5 y} - x e^{x y}$

Step 5.5.2

Factor $x'$ out of $- 4 x' e^{4 x}$ .

$x' (y e^{x y}) + x' (- 4 e^{4 x}) = - 5 e^{5 y} - x e^{x y}$

Step 5.5.3

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

$x' (y e^{x y} - 4 e^{4 x}) = - 5 e^{5 y} - x e^{x y}$

Step 5.6

Divide each term in $x' (y e^{x y} - 4 e^{4 x}) = - 5 e^{5 y} - x e^{x y}$ by $y e^{x y} - 4 e^{4 x}$ and simplify.

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Step 5.6.1

Divide each term in $x' (y e^{x y} - 4 e^{4 x}) = - 5 e^{5 y} - x e^{x y}$ by $y e^{x y} - 4 e^{4 x}$ .

$\frac{x' (y e^{x y} - 4 e^{4 x})}{y e^{x y} - 4 e^{4 x}} = \frac{- 5 e^{5 y}}{y e^{x y} - 4 e^{4 x}} + \frac{- x e^{x y}}{y e^{x y} - 4 e^{4 x}}$

Step 5.6.2

Simplify the left side.

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Step 5.6.2.1

Cancel the common factor of $y e^{x y} - 4 e^{4 x}$ .

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Step 5.6.2.1.1

Cancel the common factor.

$\frac{x' (y e^{x y} - 4 e^{4 x})}{y e^{x y} - 4 e^{4 x}} = \frac{- 5 e^{5 y}}{y e^{x y} - 4 e^{4 x}} + \frac{- x e^{x y}}{y e^{x y} - 4 e^{4 x}}$

Step 5.6.2.1.2

Divide $x'$ by $1$ .

$x' = \frac{- 5 e^{5 y}}{y e^{x y} - 4 e^{4 x}} + \frac{- x e^{x y}}{y e^{x y} - 4 e^{4 x}}$

Step 5.6.3

Simplify the right side.

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Step 5.6.3.1

Combine the numerators over the common denominator.

$x' = \frac{- 5 e^{5 y} - x e^{x y}}{y e^{x y} - 4 e^{4 x}}$

Step 5.6.3.2

Factor $- 1$ out of $- 5 e^{5 y}$ .

$x' = \frac{- (5 e^{5 y}) - x e^{x y}}{y e^{x y} - 4 e^{4 x}}$

Step 5.6.3.3

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

$x' = \frac{- (5 e^{5 y}) - (x e^{x y})}{y e^{x y} - 4 e^{4 x}}$

Step 5.6.3.4

Factor $- 1$ out of $- (5 e^{5 y}) - (x e^{x y})$ .

$x' = \frac{- (5 e^{5 y} + x e^{x y})}{y e^{x y} - 4 e^{4 x}}$

Step 5.6.3.5

Simplify the expression.

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Step 5.6.3.5.1

Rewrite $- (5 e^{5 y} + x e^{x y})$ as $- 1 (5 e^{5 y} + x e^{x y})$ .

$x' = \frac{- 1 (5 e^{5 y} + x e^{x y})}{y e^{x y} - 4 e^{4 x}}$

Step 5.6.3.5.2

Move the negative in front of the fraction.

$x' = - \frac{5 e^{5 y} + x e^{x y}}{y e^{x y} - 4 e^{4 x}}$

Step 6

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

$\frac{d x}{d y} = - \frac{5 e^{5 y} + x e^{x y}}{y e^{x y} - 4 e^{4 x}}$