Find dx/dy 5y^2-8x^4+3=0

$5 y^{2} - 8 x^{4} + 3 = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (5 y^{2} - 8 x^{4} + 3) = \frac{d}{d y} (0)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $5 y^{2} - 8 x^{4} + 3$ with respect to $y$ is $\frac{d}{d y} [5 y^{2}] + \frac{d}{d y} [- 8 x^{4}] + \frac{d}{d y} [3]$ .

$\frac{d}{d y} [5 y^{2}] + \frac{d}{d y} [- 8 x^{4}] + \frac{d}{d y} [3]$

Evaluate $\frac{d}{d y} [5 y^{2}]$ .

Since $5$ is constant with respect to $y$ , the derivative of $5 y^{2}$ with respect to $y$ is $5 \frac{d}{d y} [y^{2}]$ .

$5 \frac{d}{d y} [y^{2}] + \frac{d}{d y} [- 8 x^{4}] + \frac{d}{d y} [3]$

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

$5 (2 y) + \frac{d}{d y} [- 8 x^{4}] + \frac{d}{d y} [3]$

Multiply $2$ by $5$ .

$10 y + \frac{d}{d y} [- 8 x^{4}] + \frac{d}{d y} [3]$

Evaluate $\frac{d}{d y} [- 8 x^{4}]$ .

Since $- 8$ is constant with respect to $y$ , the derivative of $- 8 x^{4}$ with respect to $y$ is $- 8 \frac{d}{d y} [x^{4}]$ .

$10 y - 8 \frac{d}{d y} [x^{4}] + \frac{d}{d y} [3]$

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{4}$ and $g (y) = x$ .

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

$10 y - 8 (\frac{d}{d u} [u^{4}] \frac{d}{d y} [x]) + \frac{d}{d y} [3]$

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

$10 y - 8 (4 u^{3} \frac{d}{d y} [x]) + \frac{d}{d y} [3]$

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

$10 y - 8 (4 x^{3} \frac{d}{d y} [x]) + \frac{d}{d y} [3]$

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

$10 y - 8 (4 x^{3} x') + \frac{d}{d y} [3]$

Multiply $4$ by $- 8$ .

$10 y - 32 x^{3} x' + \frac{d}{d y} [3]$

Since $3$ is constant with respect to $y$ , the derivative of $3$ with respect to $y$ is $0$ .

$10 y - 32 x^{3} x' + 0$

Simplify.

Add $10 y - 32 x^{3} x'$ and $0$ .

$10 y - 32 x^{3} x'$

Reorder terms.

$- 32 x^{3} x' + 10 y$

Step 3

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

$0$

Step 4

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

$- 32 x^{3} x' + 10 y = 0$

Step 5

Solve for $x'$ .

Subtract $10 y$ from both sides of the equation.

$- 32 x^{3} x' = - 10 y$

Divide each term in $- 32 x^{3} x' = - 10 y$ by $- 32 x^{3}$ and simplify.

Divide each term in $- 32 x^{3} x' = - 10 y$ by $- 32 x^{3}$ .

$\frac{- 32 x^{3} x'}{- 32 x^{3}} = \frac{- 10 y}{- 32 x^{3}}$

Simplify the left side.

Cancel the common factor of $- 32$ .

Cancel the common factor.

$\frac{- 32 x^{3} x'}{- 32 x^{3}} = \frac{- 10 y}{- 32 x^{3}}$

Rewrite the expression.

$\frac{x^{3} x'}{x^{3}} = \frac{- 10 y}{- 32 x^{3}}$

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

Cancel the common factor.

$\frac{x^{3} x'}{x^{3}} = \frac{- 10 y}{- 32 x^{3}}$

Divide $x'$ by $1$ .

$x' = \frac{- 10 y}{- 32 x^{3}}$

Simplify the right side.

Cancel the common factor of $- 10$ and $- 32$ .

Factor $- 2$ out of $- 10 y$ .

$x' = \frac{- 2 (5 y)}{- 32 x^{3}}$

Cancel the common factors.

Factor $- 2$ out of $- 32 x^{3}$ .

$x' = \frac{- 2 (5 y)}{- 2 (16 x^{3})}$

Cancel the common factor.

$x' = \frac{- 2 (5 y)}{- 2 (16 x^{3})}$

Rewrite the expression.

$x' = \frac{5 y}{16 x^{3}}$

Step 6

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

$\frac{d x}{d y} = \frac{5 y}{16 x^{3}}$

Find dx/dy 5y^2-8x^4+3=0

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