Calculus Examples

$4 x^{2} + 2 y^{2} = 12$ , $(1, - 2)$

Step 1

Find the first derivative and evaluate at $x = 1$ and $y = - 2$ to find the slope of the tangent line.

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Step 1.1

Differentiate both sides of the equation.

$\frac{d}{d x} (4 x^{2} + 2 y^{2}) = \frac{d}{d x} (12)$

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 $4 x^{2} + 2 y^{2}$ with respect to $x$ is $\frac{d}{d x} [4 x^{2}] + \frac{d}{d x} [2 y^{2}]$ .

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

Step 1.2.2

Evaluate $\frac{d}{d x} [4 x^{2}]$ .

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Step 1.2.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{2}$ with respect to $x$ is $4 \frac{d}{d x} [x^{2}]$ .

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

Step 1.2.2.2

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

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

Step 1.2.2.3

Multiply $2$ by $4$ .

$8 x + \frac{d}{d x} [2 y^{2}]$

Step 1.2.3

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

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Step 1.2.3.1

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

$8 x + 2 \frac{d}{d x} [y^{2}]$

Step 1.2.3.2

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

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Step 1.2.3.2.1

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

$8 x + 2 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y])$

Step 1.2.3.2.2

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

$8 x + 2 (2 u \frac{d}{d x} [y])$

Step 1.2.3.2.3

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

$8 x + 2 (2 y \frac{d}{d x} [y])$

Step 1.2.3.3

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

$8 x + 2 (2 y y')$

Step 1.2.3.4

Multiply $2$ by $2$ .

$8 x + 4 y y'$

Step 1.2.4

Reorder terms.

$4 y y' + 8 x$

Step 1.3

Since $12$ is constant with respect to $x$ , the derivative of $12$ with respect to $x$ is $0$ .

$0$

Step 1.4

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

$4 y y' + 8 x = 0$

Step 1.5

Solve for $y'$ .

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Step 1.5.1

Subtract $8 x$ from both sides of the equation.

$4 y y' = - 8 x$

Step 1.5.2

Divide each term in $4 y y' = - 8 x$ by $4 y$ and simplify.

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Step 1.5.2.1

Divide each term in $4 y y' = - 8 x$ by $4 y$ .

$\frac{4 y y'}{4 y} = \frac{- 8 x}{4 y}$

Step 1.5.2.2

Simplify the left side.

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Step 1.5.2.2.1

Cancel the common factor of $4$ .

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Step 1.5.2.2.1.1

Cancel the common factor.

$\frac{4 y y'}{4 y} = \frac{- 8 x}{4 y}$

Step 1.5.2.2.1.2

Rewrite the expression.

$\frac{y y'}{y} = \frac{- 8 x}{4 y}$

Step 1.5.2.2.2

Cancel the common factor of $y$ .

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Step 1.5.2.2.2.1

Cancel the common factor.

$\frac{y y'}{y} = \frac{- 8 x}{4 y}$

Step 1.5.2.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 8 x}{4 y}$

Step 1.5.2.3

Simplify the right side.

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Step 1.5.2.3.1

Cancel the common factor of $- 8$ and $4$ .

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Step 1.5.2.3.1.1

Factor $4$ out of $- 8 x$ .

$y' = \frac{4 (- 2 x)}{4 y}$

Step 1.5.2.3.1.2

Cancel the common factors.

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Step 1.5.2.3.1.2.1

Factor $4$ out of $4 y$ .

$y' = \frac{4 (- 2 x)}{4 (y)}$

Step 1.5.2.3.1.2.2

Cancel the common factor.

$y' = \frac{4 (- 2 x)}{4 y}$

Step 1.5.2.3.1.2.3

Rewrite the expression.

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

Step 1.5.2.3.2

Move the negative in front of the fraction.

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

Step 1.6

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

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

Step 1.7

Evaluate at $x = 1$ and $y = - 2$ .

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Step 1.7.1

Replace the variable $x$ with $1$ in the expression.

$- \frac{2 (1)}{y}$

Step 1.7.2

Replace the variable $y$ with $- 2$ in the expression.

$- \frac{2 (1)}{- 2}$

Step 1.7.3

Move the negative one from the denominator of $\frac{1}{- 1}$ .

$- (- 1 \cdot 1)$

Step 1.7.4

Multiply $- (- 1 \cdot 1)$ .

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Step 1.7.4.1

Multiply $- 1$ by $1$ .

$- - 1$

Step 1.7.4.2

Multiply $- 1$ by $- 1$ .

$1$

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 $1$ and a given point $(1, - 2)$ 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 $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- 2) = 1 \cdot (x - (1))$

Step 2.2

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

$y + 2 = 1 \cdot (x - 1)$

Step 2.3

Solve for $y$ .

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Step 2.3.1

Multiply $x - 1$ by $1$ .

$y + 2 = x - 1$

Step 2.3.2

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

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Step 2.3.2.1

Subtract $2$ from both sides of the equation.

$y = x - 1 - 2$

Step 2.3.2.2

Subtract $2$ from $- 1$ .

$y = x - 3$

Step 3