Calculus Examples

$\ln (y + e^{2 x}) = x^{2} - 4$

Step 1

Differentiate both sides of the equation.

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

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 $f (x) = \ln (x)$ and $g (x) = y + e^{2 x}$ .

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Step 2.1.1

To apply the Chain Rule, set $u_{1}$ as $y + e^{2 x}$ .

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

Step 2.1.2

The derivative of $\ln (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1}}$ .

$\frac{1}{u_{1}} \frac{d}{d x} [y + e^{2 x}]$

Step 2.1.3

Replace all occurrences of $u_{1}$ with $y + e^{2 x}$ .

$\frac{1}{y + e^{2 x}} \frac{d}{d x} [y + e^{2 x}]$

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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 $g (x) = 2 x$ .

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

Replace all occurrences of $u_{2}$ with $2 x$ .

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

Step 2.5

Differentiate.

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Step 2.5.1

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 2.5.2

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

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

Step 2.5.3

Simplify the expression.

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Step 2.5.3.1

Multiply $2$ by $1$ .

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

Step 2.5.3.2

Move $2$ to the left of $e^{2 x}$ .

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

Step 2.6

Simplify.

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Step 2.6.1

Reorder the factors of $\frac{1}{y + e^{2 x}} (y' + 2 e^{2 x})$ .

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

Step 2.6.2

Multiply $y' + 2 e^{2 x}$ by $\frac{1}{y + e^{2 x}}$ .

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $x^{2} - 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4]$ .

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 4]$

Step 3.2

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

$2 x + \frac{d}{d x} [- 4]$

Step 3.3

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$2 x + 0$

Step 3.4

Add $2 x$ and $0$ .

$2 x$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Multiply both sides by $y + e^{2 x}$ .

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

Step 5.2

Simplify.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

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

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Step 5.2.1.1.1

Cancel the common factor.

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

Step 5.2.1.1.2

Rewrite the expression.

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

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Apply the distributive property.

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

Step 5.3

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

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

Step 6

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

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