Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

$\frac{d}{d x} [x y^{2}] + \frac{d}{d x} [4] + \frac{d}{d x} [y]$

Step 2.2

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

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Step 2.2.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = y^{2}$ .

$x \frac{d}{d x} [y^{2}] + y^{2} \frac{d}{d x} [x] + \frac{d}{d x} [4] + \frac{d}{d x} [y]$

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 $f (x) = x^{2}$ and $g (x) = y$ .

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Step 2.2.2.1

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

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

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 $n = 2$ .

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

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

$x (2 y y') + y^{2} \cdot 1 + \frac{d}{d x} [4] + \frac{d}{d x} [y]$

Step 2.2.5

Move $2$ to the left of $x$ .

$2 \cdot x y y' + y^{2} \cdot 1 + \frac{d}{d x} [4] + \frac{d}{d x} [y]$

Step 2.2.6

Multiply $y^{2}$ by $1$ .

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

Step 2.3

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

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

Step 2.4

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

$2 x y y' + y^{2} + 0 + y'$

Step 2.5

Simplify.

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Step 2.5.1

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

$2 x y y' + y^{2} + y'$

Step 2.5.2

Reorder terms.

$y^{2} + 2 x y y' + y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

$- (3 x^{2})$

Step 3.3

Multiply $3$ by $- 1$ .

$- 3 x^{2}$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

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

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

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

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

Step 5.2.4

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $2 x y + 1$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $y'$ by $1$ .

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

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

Step 5.3.3.2

Factor $- 1$ out of $- 3 x^{2}$ .

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

Step 5.3.3.3

Factor $- 1$ out of $- y^{2}$ .

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

Step 5.3.3.4

Factor $- 1$ out of $- (3 x^{2}) - (y^{2})$ .

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

Step 5.3.3.5

Simplify the expression.

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Step 5.3.3.5.1

Rewrite $- (3 x^{2} + y^{2})$ as $- 1 (3 x^{2} + y^{2})$ .

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

Step 5.3.3.5.2

Move the negative in front of the fraction.

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

Step 6

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

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