Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

Tap for more steps...

Step 2.1

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

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

Step 2.2

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

Tap for more steps...

Step 2.2.1

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

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

Step 2.2.2

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 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $2$ by $5$ .

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

Step 2.3

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

Tap for more steps...

Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $4$ by $- 8$ .

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

Step 2.4

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

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

Step 2.5

Simplify.

Tap for more steps...

Step 2.5.1

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

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

Step 2.5.2

Reorder terms.

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

Step 3

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

$0$

Step 4

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

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

Step 5

Solve for $y'$ .

Tap for more steps...

Step 5.1

Add $32 x^{3}$ to both sides of the equation.

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

Step 5.2

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

Tap for more steps...

Step 5.2.1

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

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

Step 5.2.2

Simplify the left side.

Tap for more steps...

Step 5.2.2.1

Cancel the common factor of $10$ .

Tap for more steps...

Step 5.2.2.1.1

Cancel the common factor.

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

Step 5.2.2.1.2

Rewrite the expression.

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

Step 5.2.2.2

Cancel the common factor of $y$ .

Tap for more steps...

Step 5.2.2.2.1

Cancel the common factor.

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

Step 5.2.2.2.2

Divide $y'$ by $1$ .

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

Step 5.2.3

Simplify the right side.

Tap for more steps...

Step 5.2.3.1

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

Tap for more steps...

Step 5.2.3.1.1

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

$y' = \frac{2 (16 x^{3})}{10 y}$

Step 5.2.3.1.2

Cancel the common factors.

Tap for more steps...

Step 5.2.3.1.2.1

Factor $2$ out of $10 y$ .

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

Step 5.2.3.1.2.2

Cancel the common factor.

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

Step 5.2.3.1.2.3

Rewrite the expression.

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

Step 6

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

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