Find dx/dy 9x^3-y^2-8=0

$9 x^{3} - y^{2} - 8 = 0$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (9 x^{3} - y^{2} - 8) = \frac{d}{d y} (0)$

Step 2

Differentiate the left side of the equation.

By the Sum Rule, the derivative of $9 x^{3} - y^{2} - 8$ with respect to $y$ is $\frac{d}{d y} [9 x^{3}] + \frac{d}{d y} [- y^{2}] + \frac{d}{d y} [- 8]$ .

$\frac{d}{d y} [9 x^{3}] + \frac{d}{d y} [- y^{2}] + \frac{d}{d y} [- 8]$

Evaluate $\frac{d}{d y} [9 x^{3}]$ .

Since $9$ is constant with respect to $y$ , the derivative of $9 x^{3}$ with respect to $y$ is $9 \frac{d}{d y} [x^{3}]$ .

$9 \frac{d}{d y} [x^{3}] + \frac{d}{d y} [- y^{2}] + \frac{d}{d y} [- 8]$

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

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

$9 (\frac{d}{d u} [u^{3}] \frac{d}{d y} [x]) + \frac{d}{d y} [- y^{2}] + \frac{d}{d y} [- 8]$

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

$9 (3 u^{2} \frac{d}{d y} [x]) + \frac{d}{d y} [- y^{2}] + \frac{d}{d y} [- 8]$

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

$9 (3 x^{2} \frac{d}{d y} [x]) + \frac{d}{d y} [- y^{2}] + \frac{d}{d y} [- 8]$

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

$9 (3 x^{2} x') + \frac{d}{d y} [- y^{2}] + \frac{d}{d y} [- 8]$

Multiply $3$ by $9$ .

$27 x^{2} x' + \frac{d}{d y} [- y^{2}] + \frac{d}{d y} [- 8]$

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

Since $- 1$ is constant with respect to $y$ , the derivative of $- y^{2}$ with respect to $y$ is $- \frac{d}{d y} [y^{2}]$ .

$27 x^{2} x' - \frac{d}{d y} [y^{2}] + \frac{d}{d y} [- 8]$

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

$27 x^{2} x' - (2 y) + \frac{d}{d y} [- 8]$

Multiply $2$ by $- 1$ .

$27 x^{2} x' - 2 y + \frac{d}{d y} [- 8]$

Differentiate using the Constant Rule.

Since $- 8$ is constant with respect to $y$ , the derivative of $- 8$ with respect to $y$ is $0$ .

$27 x^{2} x' - 2 y + 0$

Add $27 x^{2} x' - 2 y$ and $0$ .

$27 x^{2} x' - 2 y$

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.

$27 x^{2} x' - 2 y = 0$

Step 5

Solve for $x'$ .

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

$27 x^{2} x' = 2 y$

Divide each term in $27 x^{2} x' = 2 y$ by $27 x^{2}$ and simplify.

Divide each term in $27 x^{2} x' = 2 y$ by $27 x^{2}$ .

$\frac{27 x^{2} x'}{27 x^{2}} = \frac{2 y}{27 x^{2}}$

Simplify the left side.

Cancel the common factor of $27$ .

Cancel the common factor.

$\frac{27 x^{2} x'}{27 x^{2}} = \frac{2 y}{27 x^{2}}$

Rewrite the expression.

$\frac{x^{2} x'}{x^{2}} = \frac{2 y}{27 x^{2}}$

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

Cancel the common factor.

$\frac{x^{2} x'}{x^{2}} = \frac{2 y}{27 x^{2}}$

Divide $x'$ by $1$ .

$x' = \frac{2 y}{27 x^{2}}$

Step 6

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

$\frac{d x}{d y} = \frac{2 y}{27 x^{2}}$

Find dx/dy 9x^3-y^2-8=0

Related Questions

Support LCMGCF.com by disabling your adblocker.