Find dy/dx at (12,3) x^2+4y^2=36 , (12,3)

$x^{2} + 4 y^{2} = 36$ , $(12, 3)$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{2} + 4 y^{2}) = \frac{d}{d x} (36)$

Step 2

Differentiate the left side of the equation.

Differentiate.

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

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

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

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

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

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

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

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 $g (x) = y$ .

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

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

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

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

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

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

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

$2 x + 4 (2 y y')$

Multiply $2$ by $4$ .

$2 x + 8 y y'$

Reorder terms.

$8 y y' + 2 x$

Step 3

Since $36$ is constant with respect to $x$ , the derivative of $36$ with respect to $x$ is $0$ .

$0$

Step 4

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

$8 y y' + 2 x = 0$

Step 5

Solve for $y'$ .

Subtract $2 x$ from both sides of the equation.

$8 y y' = - 2 x$

Divide each term in $8 y y' = - 2 x$ by $8 y$ and simplify.

Divide each term in $8 y y' = - 2 x$ by $8 y$ .

$\frac{8 y y'}{8 y} = \frac{- 2 x}{8 y}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 y y'}{8 y} = \frac{- 2 x}{8 y}$

Rewrite the expression.

$\frac{y y'}{y} = \frac{- 2 x}{8 y}$

Cancel the common factor of $y$ .

Cancel the common factor.

$\frac{y y'}{y} = \frac{- 2 x}{8 y}$

Divide $y'$ by $1$ .

$y' = \frac{- 2 x}{8 y}$

Simplify the right side.

Cancel the common factor of $- 2$ and $8$ .

Factor $2$ out of $- 2 x$ .

$y' = \frac{2 (- x)}{8 y}$

Cancel the common factors.

Factor $2$ out of $8 y$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Step 6

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

$\frac{d y}{d x} = - \frac{x}{4 y}$

Step 7

Replace $x$ with $12$ and $y$ with $3$ in the expression.

$\frac{d y}{d x} = - \frac{12}{4 (3)}$

Step 8

Simplify the result.

Cancel the common factor of $12$ and $4$ .

Factor $4$ out of $12$ .

$- \frac{4 \cdot 3}{4 \cdot 3}$

Cancel the common factors.

Factor $4$ out of $4 \cdot 3$ .

$- \frac{4 \cdot 3}{4 (3)}$

Cancel the common factor.

$- \frac{4 \cdot 3}{4 \cdot 3}$

Rewrite the expression.

$- \frac{3}{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$- \frac{3}{3}$

Rewrite the expression.

$- 1 \cdot 1$

Multiply $- 1$ by $1$ .

$- 1$

Find dy/dx at (12,3) x^2+4y^2=36 , (12,3)

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