Calculus Examples

$y^{2} e^{2 x} - 4 y - x^{2} = 0$ , $(0, 4)$

Step 1

Find the first derivative and evaluate at $x = 0$ and $y = 4$ to find the slope of the tangent line.

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Step 1.1

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{2} e^{2 x} - 4 y - x^{2}) = \frac{d}{d x} (0)$

Step 1.2

Differentiate the left side of the equation.

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Step 1.2.1

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

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

Step 1.2.2

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

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Step 1.2.2.1

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

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

Step 1.2.2.2

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 $g (x) = 2 x$ .

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Step 1.2.2.2.1

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

$y^{2} (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [2 x]) + e^{2 x} \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [- x^{2}]$

Step 1.2.2.2.2

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

$y^{2} (e^{u_{1}} \frac{d}{d x} [2 x]) + e^{2 x} \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [- x^{2}]$

Step 1.2.2.2.3

Replace all occurrences of $u_{1}$ with $2 x$ .

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

Step 1.2.2.3

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

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

Step 1.2.2.4

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

$y^{2} (e^{2 x} (2 \cdot 1)) + e^{2 x} \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [- x^{2}]$

Step 1.2.2.5

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = y$ .

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Step 1.2.2.5.1

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

$y^{2} (e^{2 x} (2 \cdot 1)) + e^{2 x} (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [y]) + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [- x^{2}]$

Step 1.2.2.5.2

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

$y^{2} (e^{2 x} (2 \cdot 1)) + e^{2 x} (2 u_{2} \frac{d}{d x} [y]) + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [- x^{2}]$

Step 1.2.2.5.3

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

$y^{2} (e^{2 x} (2 \cdot 1)) + e^{2 x} (2 y \frac{d}{d x} [y]) + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [- x^{2}]$

Step 1.2.2.6

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

$y^{2} (e^{2 x} (2 \cdot 1)) + e^{2 x} (2 y y') + \frac{d}{d x} [- 4 y] + \frac{d}{d x} [- x^{2}]$

Step 1.2.2.7

Multiply $2$ by $1$ .

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

Step 1.2.2.8

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

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

Step 1.2.2.9

Move $2$ to the left of $y^{2}$ .

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

Step 1.2.2.10

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

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

Step 1.2.3

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

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Step 1.2.3.1

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

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

Step 1.2.3.2

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

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

Step 1.2.4

Evaluate $\frac{d}{d x} [- x^{2}]$ .

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Step 1.2.4.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

Step 1.2.4.2

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

$2 y^{2} e^{2 x} + 2 e^{2 x} y y' - 4 y' - (2 x)$

Step 1.2.4.3

Multiply $2$ by $- 1$ .

$2 y^{2} e^{2 x} + 2 e^{2 x} y y' - 4 y' - 2 x$

Step 1.2.5

Simplify.

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Step 1.2.5.1

Reorder terms.

$2 e^{2 x} y^{2} + 2 e^{2 x} y y' - 2 x - 4 y'$

Step 1.2.5.2

Reorder factors in $2 e^{2 x} y^{2} + 2 e^{2 x} y y' - 2 x - 4 y'$ .

$2 y^{2} e^{2 x} + 2 y e^{2 x} y' - 2 x - 4 y'$

Step 1.3

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

$0$

Step 1.4

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

$2 y^{2} e^{2 x} + 2 y e^{2 x} y' - 2 x - 4 y' = 0$

Step 1.5

Solve for $y'$ .

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Step 1.5.1

Reorder factors in $2 y^{2} e^{2 x} + 2 y e^{2 x} y' - 2 x - 4 y'$ .

$2 y^{2} e^{2 x} + 2 y y' e^{2 x} - 2 x - 4 y' = 0$

Step 1.5.2

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

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Step 1.5.2.1

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

$2 y y' e^{2 x} - 2 x - 4 y' = - 2 y^{2} e^{2 x}$

Step 1.5.2.2

Add $2 x$ to both sides of the equation.

$2 y y' e^{2 x} - 4 y' = - 2 y^{2} e^{2 x} + 2 x$

Step 1.5.3

Factor $2 y'$ out of $2 y y' e^{2 x} - 4 y'$ .

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Step 1.5.3.1

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

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

Step 1.5.3.2

Factor $2 y'$ out of $- 4 y'$ .

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

Step 1.5.3.3

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

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

Step 1.5.4

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

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Step 1.5.4.1

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

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

Step 1.5.4.2

Simplify the left side.

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Step 1.5.4.2.1

Cancel the common factor of $2$ .

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Step 1.5.4.2.1.1

Cancel the common factor.

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

Step 1.5.4.2.1.2

Rewrite the expression.

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

Step 1.5.4.2.2

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

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Step 1.5.4.2.2.1

Cancel the common factor.

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

Step 1.5.4.2.2.2

Divide $y'$ by $1$ .

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

Step 1.5.4.3

Simplify the right side.

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Step 1.5.4.3.1

Simplify each term.

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Step 1.5.4.3.1.1

Cancel the common factor of $- 2$ and $2$ .

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Step 1.5.4.3.1.1.1

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

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

Step 1.5.4.3.1.1.2

Cancel the common factors.

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Step 1.5.4.3.1.1.2.1

Cancel the common factor.

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

Step 1.5.4.3.1.1.2.2

Rewrite the expression.

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

Step 1.5.4.3.1.2

Move the negative in front of the fraction.

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

Step 1.5.4.3.1.3

Cancel the common factor of $2$ .

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Step 1.5.4.3.1.3.1

Cancel the common factor.

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

Step 1.5.4.3.1.3.2

Rewrite the expression.

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

Step 1.5.4.3.2

Simplify terms.

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Step 1.5.4.3.2.1

Combine the numerators over the common denominator.

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

Step 1.5.4.3.2.2

Factor $- 1$ out of $- y^{2} e^{2 x}$ .

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

Step 1.5.4.3.2.3

Factor $- 1$ out of $x$ .

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

Step 1.5.4.3.2.4

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

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

Step 1.5.4.3.2.5

Simplify the expression.

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Step 1.5.4.3.2.5.1

Rewrite $- (y^{2} e^{2 x} - x)$ as $- 1 (y^{2} e^{2 x} - x)$ .

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

Step 1.5.4.3.2.5.2

Move the negative in front of the fraction.

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

Step 1.6

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

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

Step 1.7

Evaluate at $x = 0$ and $y = 4$ .

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Step 1.7.1

Replace the variable $x$ with $0$ in the expression.

$- \frac{y^{2} e^{2 (0)} - (0)}{y e^{2 (0)} - 2}$

Step 1.7.2

Replace the variable $y$ with $4$ in the expression.

$- \frac{{(4)}^{2} e^{2 (0)} - (0)}{(4) \cdot e^{2 (0)} - 2}$

Step 1.7.3

Simplify the numerator.

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Step 1.7.3.1

Rewrite $4^{2} e^{2 (0)}$ as ${(4 e^{0})}^{2}$ .

$- \frac{{(4 e^{0})}^{2} - (0)}{4 \cdot e^{2 (0)} - 2}$

Step 1.7.3.2

Rewrite $0$ as $0^{2}$ .

$- \frac{{(4 e^{0})}^{2} - 0^{2}}{4 \cdot e^{2 (0)} - 2}$

Step 1.7.3.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4 e^{0}$ and $b = 0$ .

$- \frac{(4 e^{0} + 0) (4 e^{0} + 0)}{4 \cdot e^{2 (0)} - 2}$

Step 1.7.3.4

Simplify.

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Step 1.7.3.4.1

Anything raised to $0$ is $1$ .

$- \frac{(4 \cdot 1 + 0) (4 e^{0} + 0)}{4 \cdot e^{2 (0)} - 2}$

Step 1.7.3.4.2

Multiply $4$ by $1$ .

$- \frac{(4 + 0) (4 e^{0} + 0)}{4 \cdot e^{2 (0)} - 2}$

Step 1.7.3.4.3

Add $4$ and $0$ .

$- \frac{4 (4 e^{0} + 0)}{4 \cdot e^{2 (0)} - 2}$

Step 1.7.3.4.4

Anything raised to $0$ is $1$ .

$- \frac{4 (4 \cdot 1 + 0)}{4 \cdot e^{2 (0)} - 2}$

Step 1.7.3.4.5

Multiply $4$ by $1$ .

$- \frac{4 (4 + 0)}{4 \cdot e^{2 (0)} - 2}$

Step 1.7.3.4.6

Add $4$ and $0$ .

$- \frac{4 \cdot 4}{4 \cdot e^{2 (0)} - 2}$

Step 1.7.4

Simplify the denominator.

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Step 1.7.4.1

Multiply $2$ by $0$ .

$- \frac{4 \cdot 4}{4 \cdot e^{0} - 2}$

Step 1.7.4.2

Anything raised to $0$ is $1$ .

$- \frac{4 \cdot 4}{4 \cdot 1 - 2}$

Step 1.7.4.3

Multiply $4$ by $1$ .

$- \frac{4 \cdot 4}{4 - 2}$

Step 1.7.4.4

Subtract $2$ from $4$ .

$- \frac{4 \cdot 4}{2}$

Step 1.7.5

Simplify the expression.

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Step 1.7.5.1

Multiply $4$ by $4$ .

$- \frac{16}{2}$

Step 1.7.5.2

Divide $16$ by $2$ .

$- 1 \cdot 8$

Step 1.7.5.3

Multiply $- 1$ by $8$ .

$- 8$

Step 2

Plug the slope and point values into the point-slope formula and solve for $y$ .

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Step 2.1

Use the slope $- 8$ and a given point $(0, 4)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (4) = - 8 \cdot (x - (0))$

Step 2.2

Simplify the equation and keep it in point-slope form.

$y - 4 = - 8 \cdot (x + 0)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Add $x$ and $0$ .

$y - 4 = - 8 x$

Step 2.3.2

Add $4$ to both sides of the equation.

$y = - 8 x + 4$

Step 3