Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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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 $g (x) = y$ .

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Step 2.2.1.1

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

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

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 $n = 3$ .

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

Step 2.2.1.3

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = y$ .

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Step 2.3.1.1

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

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

Step 2.3.1.2

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

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

Step 2.3.1.3

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

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

Step 2.3.2

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

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.5

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

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Step 2.5.1

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

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

Step 2.5.2

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

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

Step 2.5.3

Multiply $2$ by $- 1$ .

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

Step 2.6

Reorder terms.

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

Step 3

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4$ with respect to $x$ is $0$ .

$0$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

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

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

Step 5.2

Factor $y'$ out of $3 y^{2} y' + 2 y y' - 5 y'$ .

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

Factor $y'$ out of $- 5 y'$ .

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

Step 5.2.4

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

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

Step 5.2.5

Factor $y'$ out of $y' (3 y^{2} + 2 y) + y' \cdot - 5$ .

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

Step 5.3

Factor.

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Step 5.3.1

Factor by grouping.

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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 $a \cdot c = 3 \cdot - 5 = - 15$ and whose sum is $b = 2$ .

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Step 5.3.1.1.1

Factor $2$ out of $2 y$ .

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

Step 5.3.1.1.2

Rewrite $2$ as $- 3$ plus $5$

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

Step 5.3.1.1.3

Apply the distributive property.

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

Step 5.3.1.2

Factor out the greatest common factor from each group.

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Step 5.3.1.2.1

Group the first two terms and the last two terms.

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

Step 5.3.1.2.2

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

$y' (3 y (y - 1) + 5 (y - 1)) = 2 x$

Step 5.3.1.3

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

$y' ((y - 1) (3 y + 5)) = 2 x$

Step 5.3.2

Remove unnecessary parentheses.

$y' (y - 1) (3 y + 5) = 2 x$

Step 5.4

Divide each term in $y' (y - 1) (3 y + 5) = 2 x$ by $(y - 1) (3 y + 5)$ and simplify.

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Step 5.4.1

Divide each term in $y' (y - 1) (3 y + 5) = 2 x$ by $(y - 1) (3 y + 5)$ .

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $y - 1$ .

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Rewrite the expression.

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

Step 5.4.2.2

Cancel the common factor of $3 y + 5$ .

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Step 5.4.2.2.1

Cancel the common factor.

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

Step 5.4.2.2.2

Divide $y'$ by $1$ .

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

Step 6

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

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