Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- y^{3}$ with respect to $x$ is $- \frac{d}{d x} [y^{3}]$ .

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

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 $f (x) = x^{3}$ and $g (x) = y$ .

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Step 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $3$ by $- 1$ .

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

Step 2.3

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$- 3 y^{2} y' + 0 + \frac{d}{d x} [- y^{2}] + \frac{d}{d x} [y]$

Step 2.4

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

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Step 2.4.1

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

$- 3 y^{2} y' + 0 - \frac{d}{d x} [y^{2}] + \frac{d}{d x} [y]$

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 $g (x) = y$ .

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Step 2.4.2.1

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

$- 3 y^{2} y' + 0 - (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [y]) + \frac{d}{d x} [y]$

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 $n = 2$ .

$- 3 y^{2} y' + 0 - (2 u_{2} \frac{d}{d x} [y]) + \frac{d}{d x} [y]$

Step 2.4.2.3

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

$- 3 y^{2} y' + 0 - (2 y \frac{d}{d x} [y]) + \frac{d}{d x} [y]$

Step 2.4.3

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

$- 3 y^{2} y' + 0 - (2 y y') + \frac{d}{d x} [y]$

Step 2.4.4

Multiply $2$ by $- 1$ .

$- 3 y^{2} y' + 0 - 2 y y' + \frac{d}{d x} [y]$

Step 2.5

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

$- 3 y^{2} y' + 0 - 2 y y' + y'$

Step 2.6

Add $- 3 y^{2} y'$ and $0$ .

$- 3 y^{2} y' - 2 y y' + y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

$2 (3 x^{2})$

Step 3.3

Multiply $3$ by $2$ .

$6 x^{2}$

Step 4

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

$- 3 y^{2} y' - 2 y y' + y' = 6 x^{2}$

Step 5

Solve for $y'$ .

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Step 5.1

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

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Step 5.1.1

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

$y' (- 3 y^{2}) - 2 y y' + y' = 6 x^{2}$

Step 5.1.2

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

$y' (- 3 y^{2}) + y' (- 2 y) + y' = 6 x^{2}$

Step 5.1.3

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

$y' (- 3 y^{2}) + y' (- 2 y) + y' = 6 x^{2}$

Step 5.1.4

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

$y' (- 3 y^{2}) + y' (- 2 y) + y' \cdot 1 = 6 x^{2}$

Step 5.1.5

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

$y' (- 3 y^{2} - 2 y) + y' \cdot 1 = 6 x^{2}$

Step 5.1.6

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

$y' (- 3 y^{2} - 2 y + 1) = 6 x^{2}$

Step 5.2

Factor.

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Step 5.2.1

Factor by grouping.

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Step 5.2.1.1

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 $b = - 2$ .

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Step 5.2.1.1.1

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

$y' (- 3 y^{2} - 2 y + 1) = 6 x^{2}$

Step 5.2.1.1.2

Rewrite $- 2$ as $1$ plus $- 3$

$y' (- 3 y^{2} + (1 - 3) y + 1) = 6 x^{2}$

Step 5.2.1.1.3

Apply the distributive property.

$y' (- 3 y^{2} + 1 y - 3 y + 1) = 6 x^{2}$

Step 5.2.1.1.4

Multiply $y$ by $1$ .

$y' (- 3 y^{2} + y - 3 y + 1) = 6 x^{2}$

Step 5.2.1.2

Factor out the greatest common factor from each group.

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Step 5.2.1.2.1

Group the first two terms and the last two terms.

$y' ((- 3 y^{2} + y) - 3 y + 1) = 6 x^{2}$

Step 5.2.1.2.2

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

$y' (y (- 3 y + 1) + 1 (- 3 y + 1)) = 6 x^{2}$

Step 5.2.1.3

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

$y' ((- 3 y + 1) (y + 1)) = 6 x^{2}$

Step 5.2.2

Remove unnecessary parentheses.

$y' (- 3 y + 1) (y + 1) = 6 x^{2}$

Step 5.3

Divide each term in $y' (- 3 y + 1) (y + 1) = 6 x^{2}$ by $(- 3 y + 1) (y + 1)$ and simplify.

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Step 5.3.1

Divide each term in $y' (- 3 y + 1) (y + 1) = 6 x^{2}$ by $(- 3 y + 1) (y + 1)$ .

$\frac{y' (- 3 y + 1) (y + 1)}{(- 3 y + 1) (y + 1)} = \frac{6 x^{2}}{(- 3 y + 1) (y + 1)}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

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

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{y' (- 3 y + 1) (y + 1)}{(- 3 y + 1) (y + 1)} = \frac{6 x^{2}}{(- 3 y + 1) (y + 1)}$

Step 5.3.2.1.2

Rewrite the expression.

$\frac{y' (y + 1)}{y + 1} = \frac{6 x^{2}}{(- 3 y + 1) (y + 1)}$

Step 5.3.2.2

Cancel the common factor of $y + 1$ .

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Step 5.3.2.2.1

Cancel the common factor.

$\frac{y' (y + 1)}{y + 1} = \frac{6 x^{2}}{(- 3 y + 1) (y + 1)}$

Step 5.3.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{6 x^{2}}{(- 3 y + 1) (y + 1)}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Factor $- 1$ out of $- 3 y$ .

$y' = \frac{6 x^{2}}{(- (3 y) + 1) (y + 1)}$

Step 5.3.3.2

Rewrite $1$ as $- 1 (- 1)$ .

$y' = \frac{6 x^{2}}{(- (3 y) - 1 (- 1)) (y + 1)}$

Step 5.3.3.3

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

$y' = \frac{6 x^{2}}{- (3 y - 1) (y + 1)}$

Step 5.3.3.4

Rewrite negatives.

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Step 5.3.3.4.1

Rewrite $- (3 y - 1)$ as $- 1 (3 y - 1)$ .

$y' = \frac{6 x^{2}}{- 1 (3 y - 1) (y + 1)}$

Step 5.3.3.4.2

Move the negative in front of the fraction.

$y' = - \frac{6 x^{2}}{(3 y - 1) (y + 1)}$

Step 6

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

$\frac{d y}{d x} = - \frac{6 x^{2}}{(3 y - 1) (y + 1)}$