Find dy/dx 7x^3-2xy+3y^2=7

$7 x^{3} - 2 x y + 3 y^{2} = 7$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (7 x^{3} - 2 x y + 3 y^{2}) = \frac{d}{d x} (7)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $7 x^{3} - 2 x y + 3 y^{2}$ with respect to $x$ is $\frac{d}{d x} [7 x^{3}] + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [3 y^{2}]$ .

$\frac{d}{d x} [7 x^{3}] + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [3 y^{2}]$

Evaluate $\frac{d}{d x} [7 x^{3}]$ .

Since $7$ is constant with respect to $x$ , the derivative of $7 x^{3}$ with respect to $x$ is $7 \frac{d}{d x} [x^{3}]$ .

$7 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [3 y^{2}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$7 (3 x^{2}) + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [3 y^{2}]$

Multiply $3$ by $7$ .

$21 x^{2} + \frac{d}{d x} [- 2 x y] + \frac{d}{d x} [3 y^{2}]$

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

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

$21 x^{2} - 2 \frac{d}{d x} [x y] + \frac{d}{d x} [3 y^{2}]$

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 $f (x) = x$ and $g (x) = y$ .

$21 x^{2} - 2 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x]) + \frac{d}{d x} [3 y^{2}]$

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

$21 x^{2} - 2 (x y' + y \frac{d}{d x} [x]) + \frac{d}{d x} [3 y^{2}]$

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

$21 x^{2} - 2 (x y' + y \cdot 1) + \frac{d}{d x} [3 y^{2}]$

Multiply $y$ by $1$ .

$21 x^{2} - 2 (x y' + y) + \frac{d}{d x} [3 y^{2}]$

Evaluate $\frac{d}{d x} [3 y^{2}]$ .

Since $3$ is constant with respect to $x$ , the derivative of $3 y^{2}$ with respect to $x$ is $3 \frac{d}{d x} [y^{2}]$ .

$21 x^{2} - 2 (x y' + y) + 3 \frac{d}{d x} [y^{2}]$

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 $y$ .

$21 x^{2} - 2 (x y' + y) + 3 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y])$

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

$21 x^{2} - 2 (x y' + y) + 3 (2 u \frac{d}{d x} [y])$

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

$21 x^{2} - 2 (x y' + y) + 3 (2 y \frac{d}{d x} [y])$

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

$21 x^{2} - 2 (x y' + y) + 3 (2 y y')$

Multiply $2$ by $3$ .

$21 x^{2} - 2 (x y' + y) + 6 y y'$

Simplify.

Apply the distributive property.

$21 x^{2} - 2 (x y') - 2 y + 6 y y'$

Remove unnecessary parentheses.

$21 x^{2} - 2 x y' - 2 y + 6 y y'$

Reorder terms.

$21 x^{2} - 2 x y' + 6 y y' - 2 y$

Step 3

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$0$

Step 4

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

$21 x^{2} - 2 x y' + 6 y y' - 2 y = 0$

Step 5

Solve for $y'$ .

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

Subtract $21 x^{2}$ from both sides of the equation.

$- 2 x y' + 6 y y' - 2 y = - 21 x^{2}$

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

$- 2 x y' + 6 y y' = - 21 x^{2} + 2 y$

Factor $2 y'$ out of $- 2 x y' + 6 y y'$ .

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

$2 y' (- x) + 6 y y' = - 21 x^{2} + 2 y$

Factor $2 y'$ out of $6 y y'$ .

$2 y' (- x) + 2 y' (3 y) = - 21 x^{2} + 2 y$

Factor $2 y'$ out of $2 y' (- x) + 2 y' (3 y)$ .

$2 y' (- x + 3 y) = - 21 x^{2} + 2 y$

Divide each term in $2 y' (- x + 3 y) = - 21 x^{2} + 2 y$ by $2 (- x + 3 y)$ and simplify.

Divide each term in $2 y' (- x + 3 y) = - 21 x^{2} + 2 y$ by $2 (- x + 3 y)$ .

$\frac{2 y' (- x + 3 y)}{2 (- x + 3 y)} = \frac{- 21 x^{2}}{2 (- x + 3 y)} + \frac{2 y}{2 (- x + 3 y)}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y' (- x + 3 y)}{2 (- x + 3 y)} = \frac{- 21 x^{2}}{2 (- x + 3 y)} + \frac{2 y}{2 (- x + 3 y)}$

Rewrite the expression.

$\frac{y' (- x + 3 y)}{- x + 3 y} = \frac{- 21 x^{2}}{2 (- x + 3 y)} + \frac{2 y}{2 (- x + 3 y)}$

Cancel the common factor of $- x + 3 y$ .

Cancel the common factor.

$\frac{y' (- x + 3 y)}{- x + 3 y} = \frac{- 21 x^{2}}{2 (- x + 3 y)} + \frac{2 y}{2 (- x + 3 y)}$

Divide $y'$ by $1$ .

$y' = \frac{- 21 x^{2}}{2 (- x + 3 y)} + \frac{2 y}{2 (- x + 3 y)}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y' = - \frac{(21) x^{2}}{2 (- x + 3 y)} + \frac{2 y}{2 (- x + 3 y)}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y' = - \frac{21 x^{2}}{2 (- x + 3 y)} + \frac{2 y}{2 (- x + 3 y)}$

Rewrite the expression.

$y' = - \frac{21 x^{2}}{2 (- x + 3 y)} + \frac{y}{- x + 3 y}$

To write $\frac{y}{- x + 3 y}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$y' = - \frac{21 x^{2}}{2 (- x + 3 y)} + \frac{y}{- x + 3 y} \cdot \frac{2}{2}$

Write each expression with a common denominator of $2 (- x + 3 y)$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{y}{- x + 3 y}$ by $\frac{2}{2}$ .

$y' = - \frac{21 x^{2}}{2 (- x + 3 y)} + \frac{y \cdot 2}{(- x + 3 y) \cdot 2}$

Reorder the factors of $(- x + 3 y) \cdot 2$ .

$y' = - \frac{21 x^{2}}{2 (- x + 3 y)} + \frac{y \cdot 2}{2 (- x + 3 y)}$

Combine the numerators over the common denominator.

$y' = \frac{- 21 x^{2} + y \cdot 2}{2 (- x + 3 y)}$

Move $2$ to the left of $y$ .

$y' = \frac{- 21 x^{2} + 2 y}{2 (- x + 3 y)}$

Simplify terms.

Factor $- 1$ out of $- 21 x^{2}$ .

$y' = \frac{- (21 x^{2}) + 2 y}{2 (- x + 3 y)}$

Factor $- 1$ out of $2 y$ .

$y' = \frac{- (21 x^{2}) - (- 2 y)}{2 (- x + 3 y)}$

Factor $- 1$ out of $- (21 x^{2}) - (- 2 y)$ .

$y' = \frac{- (21 x^{2} - 2 y)}{2 (- x + 3 y)}$

Rewrite $- (21 x^{2} - 2 y)$ as $- 1 (21 x^{2} - 2 y)$ .

$y' = \frac{- 1 (21 x^{2} - 2 y)}{2 (- x + 3 y)}$

Factor $- 1$ out of $- x$ .

$y' = \frac{- 1 (21 x^{2} - 2 y)}{2 (- (x) + 3 y)}$

Factor $- 1$ out of $3 y$ .

$y' = \frac{- 1 (21 x^{2} - 2 y)}{2 (- (x) - (- 3 y))}$

Factor $- 1$ out of $- (x) - (- 3 y)$ .

$y' = \frac{- 1 (21 x^{2} - 2 y)}{2 (- (x - 3 y))}$

Rewrite $- (x - 3 y)$ as $- 1 (x - 3 y)$ .

$y' = \frac{- 1 (21 x^{2} - 2 y)}{2 (- 1 (x - 3 y))}$

Cancel the common factor.

$y' = \frac{- 1 (21 x^{2} - 2 y)}{2 (- 1 (x - 3 y))}$

Rewrite the expression.

$y' = \frac{21 x^{2} - 2 y}{2 (x - 3 y)}$

Step 6

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

$\frac{d y}{d x} = \frac{21 x^{2} - 2 y}{2 (x - 3 y)}$

Find dy/dx 7x^3-2xy+3y^2=7

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