Find dy/dx 6x^4=4y^2+3x^2

$6 x^{4} = 4 y^{2} + 3 x^{2}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (6 x^{4}) = \frac{d}{d x} (4 y^{2} + 3 x^{2})$

Step 2

Differentiate the left side of the equation.

Since $6$ is constant with respect to $x$ , the derivative of $6 x^{4}$ with respect to $x$ is $6 \frac{d}{d x} [x^{4}]$ .

$6 \frac{d}{d x} [x^{4}]$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$6 (4 x^{3})$

Multiply $4$ by $6$ .

$24 x^{3}$

Step 3

Differentiate the right side of the equation.

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

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

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

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

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 $y$ .

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

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

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

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

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

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

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

Multiply $2$ by $4$ .

$8 y y' + \frac{d}{d x} [3 x^{2}]$

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

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

$8 y y' + 3 \frac{d}{d x} [x^{2}]$

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

$8 y y' + 3 (2 x)$

Multiply $2$ by $3$ .

$8 y y' + 6 x$

Step 4

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

$24 x^{3} = 8 y y' + 6 x$

Step 5

Solve for $y'$ .

Rewrite the equation as $8 y y' + 6 x = 24 x^{3}$ .

$8 y y' + 6 x = 24 x^{3}$

Subtract $6 x$ from both sides of the equation.

$8 y y' = 24 x^{3} - 6 x$

Divide each term in $8 y y' = 24 x^{3} - 6 x$ by $8 y$ and simplify.

Divide each term in $8 y y' = 24 x^{3} - 6 x$ by $8 y$ .

$\frac{8 y y'}{8 y} = \frac{24 x^{3}}{8 y} + \frac{- 6 x}{8 y}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 y y'}{8 y} = \frac{24 x^{3}}{8 y} + \frac{- 6 x}{8 y}$

Rewrite the expression.

$\frac{y y'}{y} = \frac{24 x^{3}}{8 y} + \frac{- 6 x}{8 y}$

Cancel the common factor of $y$ .

Cancel the common factor.

$\frac{y y'}{y} = \frac{24 x^{3}}{8 y} + \frac{- 6 x}{8 y}$

Divide $y'$ by $1$ .

$y' = \frac{24 x^{3}}{8 y} + \frac{- 6 x}{8 y}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $24$ and $8$ .

Factor $8$ out of $24 x^{3}$ .

$y' = \frac{8 (3 x^{3})}{8 y} + \frac{- 6 x}{8 y}$

Cancel the common factors.

Factor $8$ out of $8 y$ .

$y' = \frac{8 (3 x^{3})}{8 (y)} + \frac{- 6 x}{8 y}$

Cancel the common factor.

$y' = \frac{8 (3 x^{3})}{8 y} + \frac{- 6 x}{8 y}$

Rewrite the expression.

$y' = \frac{3 x^{3}}{y} + \frac{- 6 x}{8 y}$

Cancel the common factor of $- 6$ and $8$ .

Factor $2$ out of $- 6 x$ .

$y' = \frac{3 x^{3}}{y} + \frac{2 (- 3 x)}{8 y}$

Cancel the common factors.

Factor $2$ out of $8 y$ .

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

Cancel the common factor.

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

Rewrite the expression.

$y' = \frac{3 x^{3}}{y} + \frac{- 3 x}{4 y}$

Move the negative in front of the fraction.

$y' = \frac{3 x^{3}}{y} - \frac{3 x}{4 y}$

Step 6

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

$\frac{d y}{d x} = \frac{3 x^{3}}{y} - \frac{3 x}{4 y}$

Find dy/dx 6x^4=4y^2+3x^2

Related Questions

Support LCMGCF.com by disabling your adblocker.