Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

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

$2 x$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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 $g (x) = x - y$ .

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Step 3.1.1

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

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

Step 3.1.2

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

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

Step 3.1.3

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

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

Step 3.2

Differentiate.

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.3

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

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

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Rewrite the equation as $e^{x - y} (1 - y') = 2 x$ .

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

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

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Step 5.2.2.1.1

Cancel the common factor.

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

Step 5.2.2.1.2

Divide $1 - y'$ by $1$ .

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

Step 5.3

Subtract $1$ from both sides of the equation.

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

Step 5.4

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

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Step 5.4.1

Divide each term in $- y' = \frac{2 x}{e^{x - y}} - 1$ by $- 1$ .

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Dividing two negative values results in a positive value.

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

Step 5.4.2.2

Divide $y'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Simplify each term.

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Step 5.4.3.1.1

Move the negative one from the denominator of $\frac{\frac{2 x}{e^{x - y}}}{- 1}$ .

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

Step 5.4.3.1.2

Rewrite $- 1 \cdot \frac{2 x}{e^{x - y}}$ as $- \frac{2 x}{e^{x - y}}$ .

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

Step 5.4.3.1.3

Divide $- 1$ by $- 1$ .

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

Step 6

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

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