Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

Tap for more steps...

Step 2.1

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

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

Step 2.2

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

Tap for more steps...

Step 2.2.1

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 2.2.1.1

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

$\frac{d}{d u_{1}} [{u_{1}}^{3}] \frac{d}{d x} [y] + \frac{d}{d x} [y] + \frac{d}{d x} [- 2 y^{2}] + \frac{d}{d x} [5 x^{2}]$

Step 2.2.1.2

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

$3 {u_{1}}^{2} \frac{d}{d x} [y] + \frac{d}{d x} [y] + \frac{d}{d x} [- 2 y^{2}] + \frac{d}{d x} [5 x^{2}]$

Step 2.2.1.3

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

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

Step 2.2.2

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

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

Step 2.3

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

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

Step 2.4

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

Tap for more steps...

Step 2.4.1

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

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

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 $y$ .

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

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 $n = 2$ .

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

Step 2.4.2.3

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

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

Step 2.4.3

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

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

Step 2.4.4

Multiply $2$ by $- 2$ .

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

Step 2.5

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

Tap for more steps...

Step 2.5.1

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

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

Step 2.5.2

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

$3 y^{2} y' + y' - 4 y y' + 5 (2 x)$

Step 2.5.3

Multiply $2$ by $5$ .

$3 y^{2} y' + y' - 4 y y' + 10 x$

Step 2.6

Reorder terms.

$3 y^{2} y' - 4 y y' + 10 x + y'$

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.

$3 y^{2} y' - 4 y y' + 10 x + y' = 0$

Step 5

Solve for $y'$ .

Tap for more steps...

Step 5.1

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

$3 y^{2} y' - 4 y y' + y' = - 10 x$

Step 5.2

Factor $y'$ out of $3 y^{2} y' - 4 y y' + y'$ .

Tap for more steps...

Step 5.2.1

Factor $y'$ out of $3 y^{2} y'$ .

$y' (3 y^{2}) - 4 y y' + y' = - 10 x$

Step 5.2.2

Factor $y'$ out of $- 4 y y'$ .

$y' (3 y^{2}) + y' (- 4 y) + y' = - 10 x$

Step 5.2.3

Raise $y'$ to the power of $1$ .

$y' (3 y^{2}) + y' (- 4 y) + y' = - 10 x$

Step 5.2.4

Factor $y'$ out of ${y'}^{1}$ .

$y' (3 y^{2}) + y' (- 4 y) + y' \cdot 1 = - 10 x$

Step 5.2.5

Factor $y'$ out of $y' (3 y^{2}) + y' (- 4 y)$ .

$y' (3 y^{2} - 4 y) + y' \cdot 1 = - 10 x$

Step 5.2.6

Factor $y'$ out of $y' (3 y^{2} - 4 y) + y' \cdot 1$ .

$y' (3 y^{2} - 4 y + 1) = - 10 x$

Step 5.3

Factor.

Tap for more steps...

Step 5.3.1

Factor by grouping.

Tap for more steps...

Step 5.3.1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot 1 = 3$ and whose sum is $b = - 4$ .

Tap for more steps...

Step 5.3.1.1.1

Factor $- 4$ out of $- 4 y$ .

$y' (3 y^{2} - 4 y + 1) = - 10 x$

Step 5.3.1.1.2

Rewrite $- 4$ as $- 1$ plus $- 3$

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

Step 5.3.1.1.3

Apply the distributive property.

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

Step 5.3.1.2

Factor out the greatest common factor from each group.

Tap for more steps...

Step 5.3.1.2.1

Group the first two terms and the last two terms.

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

Step 5.3.1.2.2

Factor out the greatest common factor (GCF) from each group.

$y' (y (3 y - 1) - (3 y - 1)) = - 10 x$

Step 5.3.1.3

Factor the polynomial by factoring out the greatest common factor, $3 y - 1$ .

$y' ((3 y - 1) (y - 1)) = - 10 x$

Step 5.3.2

Remove unnecessary parentheses.

$y' (3 y - 1) (y - 1) = - 10 x$

Step 5.4

Divide each term in $y' (3 y - 1) (y - 1) = - 10 x$ by $(3 y - 1) (y - 1)$ and simplify.

Tap for more steps...

Step 5.4.1

Divide each term in $y' (3 y - 1) (y - 1) = - 10 x$ by $(3 y - 1) (y - 1)$ .

$\frac{y' (3 y - 1) (y - 1)}{(3 y - 1) (y - 1)} = \frac{- 10 x}{(3 y - 1) (y - 1)}$

Step 5.4.2

Simplify the left side.

Tap for more steps...

Step 5.4.2.1

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

Tap for more steps...

Step 5.4.2.1.1

Cancel the common factor.

$\frac{y' (3 y - 1) (y - 1)}{(3 y - 1) (y - 1)} = \frac{- 10 x}{(3 y - 1) (y - 1)}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{y' (y - 1)}{y - 1} = \frac{- 10 x}{(3 y - 1) (y - 1)}$

Step 5.4.2.2

Cancel the common factor of $y - 1$ .

Tap for more steps...

Step 5.4.2.2.1

Cancel the common factor.

$\frac{y' (y - 1)}{y - 1} = \frac{- 10 x}{(3 y - 1) (y - 1)}$

Step 5.4.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 10 x}{(3 y - 1) (y - 1)}$

Step 5.4.3

Simplify the right side.

Tap for more steps...

Step 5.4.3.1

Move the negative in front of the fraction.

$y' = - \frac{10 x}{(3 y - 1) (y - 1)}$

Step 6

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

$\frac{d y}{d x} = - \frac{10 x}{(3 y - 1) (y - 1)}$