Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

Tap for more steps...

Step 2.1

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

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

Step 2.2

Evaluate $\frac{d}{d x} [5 x^{3}]$ .

Tap for more steps...

Step 2.2.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{3}$ with respect to $x$ is $5 \frac{d}{d x} [x^{3}]$ .

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $3$ by $5$ .

$15 x^{2} + \frac{d}{d x} [2 y^{3}] + \frac{d}{d x} [- y^{2}] + \frac{d}{d x} [3]$

Step 2.3

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

Tap for more steps...

Step 2.3.1

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

$15 x^{2} + 2 \frac{d}{d x} [y^{3}] + \frac{d}{d x} [- y^{2}] + \frac{d}{d x} [3]$

Step 2.3.2

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

Tap for more steps...

Step 2.3.2.1

To apply the Chain Rule, set $u_{1}$ as $y$ .

$15 x^{2} + 2 (\frac{d}{d u_{1}} [{u_{1}}^{3}] \frac{d}{d x} [y]) + \frac{d}{d x} [- y^{2}] + \frac{d}{d x} [3]$

Step 2.3.2.2

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

$15 x^{2} + 2 (3 {u_{1}}^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [- y^{2}] + \frac{d}{d x} [3]$

Step 2.3.2.3

Replace all occurrences of $u_{1}$ with $y$ .

$15 x^{2} + 2 (3 y^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [- y^{2}] + \frac{d}{d x} [3]$

Step 2.3.3

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

$15 x^{2} + 2 (3 y^{2} y') + \frac{d}{d x} [- y^{2}] + \frac{d}{d x} [3]$

Step 2.3.4

Multiply $3$ by $2$ .

$15 x^{2} + 6 y^{2} y' + \frac{d}{d x} [- y^{2}] + \frac{d}{d x} [3]$

Step 2.4

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

Tap for more steps...

Step 2.4.1

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

$15 x^{2} + 6 y^{2} y' - \frac{d}{d x} [y^{2}] + \frac{d}{d x} [3]$

Step 2.4.2

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 $g (x) = y$ .

Tap for more steps...

Step 2.4.2.1

To apply the Chain Rule, set $u_{2}$ as $y$ .

$15 x^{2} + 6 y^{2} y' - (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [y]) + \frac{d}{d x} [3]$

Step 2.4.2.2

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

$15 x^{2} + 6 y^{2} y' - (2 u_{2} \frac{d}{d x} [y]) + \frac{d}{d x} [3]$

Step 2.4.2.3

Replace all occurrences of $u_{2}$ with $y$ .

$15 x^{2} + 6 y^{2} y' - (2 y \frac{d}{d x} [y]) + \frac{d}{d x} [3]$

Step 2.4.3

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

$15 x^{2} + 6 y^{2} y' - (2 y y') + \frac{d}{d x} [3]$

Step 2.4.4

Multiply $2$ by $- 1$ .

$15 x^{2} + 6 y^{2} y' - 2 y y' + \frac{d}{d x} [3]$

Step 2.5

Differentiate using the Constant Rule.

Tap for more steps...

Step 2.5.1

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

$15 x^{2} + 6 y^{2} y' - 2 y y' + 0$

Step 2.5.2

Add $15 x^{2} + 6 y^{2} y' - 2 y y'$ and $0$ .

$15 x^{2} + 6 y^{2} y' - 2 y y'$

Step 3

Differentiate the right side of the equation.

Tap for more steps...

Step 3.1

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

$- 5 \frac{d}{d x} [x^{2}]$

Step 3.2

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

$- 5 (2 x)$

Step 3.3

Multiply $2$ by $- 5$ .

$- 10 x$

Step 4

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

$15 x^{2} + 6 y^{2} y' - 2 y y' = - 10 x$

Step 5

Solve for $y'$ .

Tap for more steps...

Step 5.1

Subtract $15 x^{2}$ from both sides of the equation.

$6 y^{2} y' - 2 y y' = - 10 x - 15 x^{2}$

Step 5.2

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

Tap for more steps...

Step 5.2.1

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

$2 y y' (3 y) - 2 y y' = - 10 x - 15 x^{2}$

Step 5.2.2

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

$2 y y' (3 y) + 2 y y' \cdot - 1 = - 10 x - 15 x^{2}$

Step 5.2.3

Factor $2 y y'$ out of $2 y y' (3 y) + 2 y y' \cdot - 1$ .

$2 y y' (3 y - 1) = - 10 x - 15 x^{2}$

Step 5.3

Divide each term in $2 y y' (3 y - 1) = - 10 x - 15 x^{2}$ by $2 y (3 y - 1)$ and simplify.

Tap for more steps...

Step 5.3.1

Divide each term in $2 y y' (3 y - 1) = - 10 x - 15 x^{2}$ by $2 y (3 y - 1)$ .

$\frac{2 y y' (3 y - 1)}{2 y (3 y - 1)} = \frac{- 10 x}{2 y (3 y - 1)} + \frac{- 15 x^{2}}{2 y (3 y - 1)}$

Step 5.3.2

Simplify the left side.

Tap for more steps...

Step 5.3.2.1

Cancel the common factor of $2$ .

Tap for more steps...

Step 5.3.2.1.1

Cancel the common factor.

$\frac{2 y y' (3 y - 1)}{2 y (3 y - 1)} = \frac{- 10 x}{2 y (3 y - 1)} + \frac{- 15 x^{2}}{2 y (3 y - 1)}$

Step 5.3.2.1.2

Rewrite the expression.

$\frac{y y' (3 y - 1)}{y (3 y - 1)} = \frac{- 10 x}{2 y (3 y - 1)} + \frac{- 15 x^{2}}{2 y (3 y - 1)}$

Step 5.3.2.2

Cancel the common factor of $y$ .

Tap for more steps...

Step 5.3.2.2.1

Cancel the common factor.

$\frac{y y' (3 y - 1)}{y (3 y - 1)} = \frac{- 10 x}{2 y (3 y - 1)} + \frac{- 15 x^{2}}{2 y (3 y - 1)}$

Step 5.3.2.2.2

Rewrite the expression.

$\frac{y' (3 y - 1)}{3 y - 1} = \frac{- 10 x}{2 y (3 y - 1)} + \frac{- 15 x^{2}}{2 y (3 y - 1)}$

Step 5.3.2.3

Cancel the common factor of $3 y - 1$ .

Tap for more steps...

Step 5.3.2.3.1

Cancel the common factor.

$\frac{y' (3 y - 1)}{3 y - 1} = \frac{- 10 x}{2 y (3 y - 1)} + \frac{- 15 x^{2}}{2 y (3 y - 1)}$

Step 5.3.2.3.2

Divide $y'$ by $1$ .

$y' = \frac{- 10 x}{2 y (3 y - 1)} + \frac{- 15 x^{2}}{2 y (3 y - 1)}$

Step 5.3.3

Simplify the right side.

Tap for more steps...

Step 5.3.3.1

Simplify each term.

Tap for more steps...

Step 5.3.3.1.1

Cancel the common factor of $- 10$ and $2$ .

Tap for more steps...

Step 5.3.3.1.1.1

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

$y' = \frac{2 (- 5 x)}{2 y (3 y - 1)} + \frac{- 15 x^{2}}{2 y (3 y - 1)}$

Step 5.3.3.1.1.2

Cancel the common factors.

Tap for more steps...

Step 5.3.3.1.1.2.1

Factor $2$ out of $2 y (3 y - 1)$ .

$y' = \frac{2 (- 5 x)}{2 (y (3 y - 1))} + \frac{- 15 x^{2}}{2 y (3 y - 1)}$

Step 5.3.3.1.1.2.2

Cancel the common factor.

$y' = \frac{2 (- 5 x)}{2 (y (3 y - 1))} + \frac{- 15 x^{2}}{2 y (3 y - 1)}$

Step 5.3.3.1.1.2.3

Rewrite the expression.

$y' = \frac{- 5 x}{y (3 y - 1)} + \frac{- 15 x^{2}}{2 y (3 y - 1)}$

Step 5.3.3.1.2

Move the negative in front of the fraction.

$y' = - \frac{5 x}{y (3 y - 1)} + \frac{- 15 x^{2}}{2 y (3 y - 1)}$

Step 5.3.3.1.3

Move the negative in front of the fraction.

$y' = - \frac{5 x}{y (3 y - 1)} - \frac{15 x^{2}}{2 y (3 y - 1)}$

Step 6

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

$\frac{d y}{d x} = - \frac{5 x}{y (3 y - 1)} - \frac{15 x^{2}}{2 y (3 y - 1)}$