Find dy/dx 3x^3+4xy+2y^3=34

$3 x^{3} + 4 x y + 2 y^{3} = 34$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (3 x^{3} + 4 x y + 2 y^{3}) = \frac{d}{d x} (34)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $3 x^{3} + 4 x y + 2 y^{3}$ with respect to $x$ is $\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [4 x y] + \frac{d}{d x} [2 y^{3}]$ .

$\frac{d}{d x} [3 x^{3}] + \frac{d}{d x} [4 x y] + \frac{d}{d x} [2 y^{3}]$

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

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

$3 \frac{d}{d x} [x^{3}] + \frac{d}{d x} [4 x y] + \frac{d}{d x} [2 y^{3}]$

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

$3 (3 x^{2}) + \frac{d}{d x} [4 x y] + \frac{d}{d x} [2 y^{3}]$

Multiply $3$ by $3$ .

$9 x^{2} + \frac{d}{d x} [4 x y] + \frac{d}{d x} [2 y^{3}]$

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

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

$9 x^{2} + 4 \frac{d}{d x} [x y] + \frac{d}{d x} [2 y^{3}]$

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 $g (x) = y$ .

$9 x^{2} + 4 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x]) + \frac{d}{d x} [2 y^{3}]$

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

$9 x^{2} + 4 (x y' + y \frac{d}{d x} [x]) + \frac{d}{d x} [2 y^{3}]$

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

$9 x^{2} + 4 (x y' + y \cdot 1) + \frac{d}{d x} [2 y^{3}]$

Multiply $y$ by $1$ .

$9 x^{2} + 4 (x y' + y) + \frac{d}{d x} [2 y^{3}]$

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

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

$9 x^{2} + 4 (x y' + y) + 2 \frac{d}{d x} [y^{3}]$

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = y$ .

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

$9 x^{2} + 4 (x y' + y) + 2 (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y])$

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

$9 x^{2} + 4 (x y' + y) + 2 (3 u^{2} \frac{d}{d x} [y])$

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

$9 x^{2} + 4 (x y' + y) + 2 (3 y^{2} \frac{d}{d x} [y])$

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

$9 x^{2} + 4 (x y' + y) + 2 (3 y^{2} y')$

Multiply $3$ by $2$ .

$9 x^{2} + 4 (x y' + y) + 6 y^{2} y'$

Simplify.

Apply the distributive property.

$9 x^{2} + 4 (x y') + 4 y + 6 y^{2} y'$

Remove unnecessary parentheses.

$9 x^{2} + 4 x y' + 4 y + 6 y^{2} y'$

Reorder terms.

$9 x^{2} + 4 x y' + 6 y^{2} y' + 4 y$

Step 3

Since $34$ is constant with respect to $x$ , the derivative of $34$ with respect to $x$ is $0$ .

$0$

Step 4

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

$9 x^{2} + 4 x y' + 6 y^{2} y' + 4 y = 0$

Step 5

Solve for $y'$ .

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

Subtract $9 x^{2}$ from both sides of the equation.

$4 x y' + 6 y^{2} y' + 4 y = - 9 x^{2}$

Subtract $4 y$ from both sides of the equation.

$4 x y' + 6 y^{2} y' = - 9 x^{2} - 4 y$

Factor $2 y'$ out of $4 x y' + 6 y^{2} y'$ .

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

$2 y' (2 x) + 6 y^{2} y' = - 9 x^{2} - 4 y$

Factor $2 y'$ out of $6 y^{2} y'$ .

$2 y' (2 x) + 2 y' (3 y^{2}) = - 9 x^{2} - 4 y$

Factor $2 y'$ out of $2 y' (2 x) + 2 y' (3 y^{2})$ .

$2 y' (2 x + 3 y^{2}) = - 9 x^{2} - 4 y$

Divide each term in $2 y' (2 x + 3 y^{2}) = - 9 x^{2} - 4 y$ by $2 (2 x + 3 y^{2})$ and simplify.

Divide each term in $2 y' (2 x + 3 y^{2}) = - 9 x^{2} - 4 y$ by $2 (2 x + 3 y^{2})$ .

$\frac{2 y' (2 x + 3 y^{2})}{2 (2 x + 3 y^{2})} = \frac{- 9 x^{2}}{2 (2 x + 3 y^{2})} + \frac{- 4 y}{2 (2 x + 3 y^{2})}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 y' (2 x + 3 y^{2})}{2 (2 x + 3 y^{2})} = \frac{- 9 x^{2}}{2 (2 x + 3 y^{2})} + \frac{- 4 y}{2 (2 x + 3 y^{2})}$

Rewrite the expression.

$\frac{y' (2 x + 3 y^{2})}{2 x + 3 y^{2}} = \frac{- 9 x^{2}}{2 (2 x + 3 y^{2})} + \frac{- 4 y}{2 (2 x + 3 y^{2})}$

Cancel the common factor of $2 x + 3 y^{2}$ .

Cancel the common factor.

$\frac{y' (2 x + 3 y^{2})}{2 x + 3 y^{2}} = \frac{- 9 x^{2}}{2 (2 x + 3 y^{2})} + \frac{- 4 y}{2 (2 x + 3 y^{2})}$

Divide $y'$ by $1$ .

$y' = \frac{- 9 x^{2}}{2 (2 x + 3 y^{2})} + \frac{- 4 y}{2 (2 x + 3 y^{2})}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y' = - \frac{(9) x^{2}}{2 (2 x + 3 y^{2})} + \frac{- 4 y}{2 (2 x + 3 y^{2})}$

Cancel the common factor of $- 4$ and $2$ .

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

$y' = - \frac{9 x^{2}}{2 (2 x + 3 y^{2})} + \frac{2 (- 2 y)}{2 (2 x + 3 y^{2})}$

Cancel the common factors.

Cancel the common factor.

$y' = - \frac{9 x^{2}}{2 (2 x + 3 y^{2})} + \frac{2 (- 2 y)}{2 (2 x + 3 y^{2})}$

Rewrite the expression.

$y' = - \frac{9 x^{2}}{2 (2 x + 3 y^{2})} + \frac{- 2 y}{2 x + 3 y^{2}}$

Move the negative in front of the fraction.

$y' = - \frac{9 x^{2}}{2 (2 x + 3 y^{2})} - \frac{2 y}{2 x + 3 y^{2}}$

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

$y' = - \frac{9 x^{2}}{2 (2 x + 3 y^{2})} - \frac{2 y}{2 x + 3 y^{2}} \cdot \frac{2}{2}$

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

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

$y' = - \frac{9 x^{2}}{2 (2 x + 3 y^{2})} - \frac{2 y \cdot 2}{(2 x + 3 y^{2}) \cdot 2}$

Reorder the factors of $(2 x + 3 y^{2}) \cdot 2$ .

$y' = - \frac{9 x^{2}}{2 (2 x + 3 y^{2})} - \frac{2 y \cdot 2}{2 (2 x + 3 y^{2})}$

Combine the numerators over the common denominator.

$y' = \frac{- 9 x^{2} - 2 y \cdot 2}{2 (2 x + 3 y^{2})}$

Multiply $2$ by $- 2$ .

$y' = \frac{- 9 x^{2} - 4 y}{2 (2 x + 3 y^{2})}$

Factor $- 1$ out of $- 9 x^{2}$ .

$y' = \frac{- (9 x^{2}) - 4 y}{2 (2 x + 3 y^{2})}$

Factor $- 1$ out of $- 4 y$ .

$y' = \frac{- (9 x^{2}) - (4 y)}{2 (2 x + 3 y^{2})}$

Factor $- 1$ out of $- (9 x^{2}) - (4 y)$ .

$y' = \frac{- (9 x^{2} + 4 y)}{2 (2 x + 3 y^{2})}$

Simplify the expression.

Rewrite $- (9 x^{2} + 4 y)$ as $- 1 (9 x^{2} + 4 y)$ .

$y' = \frac{- 1 (9 x^{2} + 4 y)}{2 (2 x + 3 y^{2})}$

Move the negative in front of the fraction.

$y' = - \frac{9 x^{2} + 4 y}{2 (2 x + 3 y^{2})}$

Step 6

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

$\frac{d y}{d x} = - \frac{9 x^{2} + 4 y}{2 (2 x + 3 y^{2})}$

Find dy/dx 3x^3+4xy+2y^3=34

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