Calculus Examples

$2 {(x^{2} + 1)}^{4} + {(y^{2} + 3)}^{2} = 24$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (2 {(x^{2} + 1)}^{4} + {(y^{2} + 3)}^{2}) = \frac{d}{d x} (24)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

$\frac{d}{d x} [2 {(x^{2} + 1)}^{4} + (y^{2} + 3) (y^{2} + 3)]$

Step 2.2

Expand $(y^{2} + 3) (y^{2} + 3)$ using the FOIL Method.

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Step 2.2.1

Apply the distributive property.

$\frac{d}{d x} [2 {(x^{2} + 1)}^{4} + y^{2} (y^{2} + 3) + 3 (y^{2} + 3)]$

Step 2.2.2

Apply the distributive property.

$\frac{d}{d x} [2 {(x^{2} + 1)}^{4} + y^{2} y^{2} + y^{2} \cdot 3 + 3 (y^{2} + 3)]$

Step 2.2.3

Apply the distributive property.

$\frac{d}{d x} [2 {(x^{2} + 1)}^{4} + y^{2} y^{2} + y^{2} \cdot 3 + 3 y^{2} + 3 \cdot 3]$

Step 2.3

Simplify and combine like terms.

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Step 2.3.1

Simplify each term.

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Step 2.3.1.1

Multiply $y^{2}$ by $y^{2}$ by adding the exponents.

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Step 2.3.1.1.1

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{d}{d x} [2 {(x^{2} + 1)}^{4} + y^{2 + 2} + y^{2} \cdot 3 + 3 y^{2} + 3 \cdot 3]$

Step 2.3.1.1.2

Add $2$ and $2$ .

$\frac{d}{d x} [2 {(x^{2} + 1)}^{4} + y^{4} + y^{2} \cdot 3 + 3 y^{2} + 3 \cdot 3]$

Step 2.3.1.2

Move $3$ to the left of $y^{2}$ .

$\frac{d}{d x} [2 {(x^{2} + 1)}^{4} + y^{4} + 3 \cdot y^{2} + 3 y^{2} + 3 \cdot 3]$

Step 2.3.1.3

Multiply $3$ by $3$ .

$\frac{d}{d x} [2 {(x^{2} + 1)}^{4} + y^{4} + 3 y^{2} + 3 y^{2} + 9]$

Step 2.3.2

Add $3 y^{2}$ and $3 y^{2}$ .

$\frac{d}{d x} [2 {(x^{2} + 1)}^{4} + y^{4} + 6 y^{2} + 9]$

Step 2.4

By the Sum Rule, the derivative of $2 {(x^{2} + 1)}^{4} + y^{4} + 6 y^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [2 {(x^{2} + 1)}^{4}] + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$ .

$\frac{d}{d x} [2 {(x^{2} + 1)}^{4}] + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.5

Evaluate $\frac{d}{d x} [2 {(x^{2} + 1)}^{4}]$ .

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Step 2.5.1

Since $2$ is constant with respect to $x$ , the derivative of $2 {(x^{2} + 1)}^{4}$ with respect to $x$ is $2 \frac{d}{d x} [{(x^{2} + 1)}^{4}]$ .

$2 \frac{d}{d x} [{(x^{2} + 1)}^{4}] + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.5.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = x^{2} + 1$ .

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Step 2.5.2.1

To apply the Chain Rule, set $u_{1}$ as $x^{2} + 1$ .

$2 (\frac{d}{d u_{1}} [{u_{1}}^{4}] \frac{d}{d x} [x^{2} + 1]) + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.5.2.2

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

$2 (4 {u_{1}}^{3} \frac{d}{d x} [x^{2} + 1]) + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.5.2.3

Replace all occurrences of $u_{1}$ with $x^{2} + 1$ .

$2 (4 {(x^{2} + 1)}^{3} \frac{d}{d x} [x^{2} + 1]) + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.5.3

By the Sum Rule, the derivative of $x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1]$ .

$2 (4 {(x^{2} + 1)}^{3} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [1])) + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.5.4

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

$2 (4 {(x^{2} + 1)}^{3} (2 x + \frac{d}{d x} [1])) + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.5.5

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$2 (4 {(x^{2} + 1)}^{3} (2 x + 0)) + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.5.6

Add $2 x$ and $0$ .

$2 (4 {(x^{2} + 1)}^{3} (2 x)) + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.5.7

Multiply $2$ by $4$ .

$2 (8 {(x^{2} + 1)}^{3} x) + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.5.8

Multiply $8$ by $2$ .

$16 {(x^{2} + 1)}^{3} x + \frac{d}{d x} [y^{4}] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.6

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

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Step 2.6.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{4}$ and $g (x) = y$ .

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Step 2.6.1.1

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

$16 {(x^{2} + 1)}^{3} x + \frac{d}{d u_{2}} [{u_{2}}^{4}] \frac{d}{d x} [y] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.6.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 4$ .

$16 {(x^{2} + 1)}^{3} x + 4 {u_{2}}^{3} \frac{d}{d x} [y] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.6.1.3

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

$16 {(x^{2} + 1)}^{3} x + 4 y^{3} \frac{d}{d x} [y] + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.6.2

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

$16 {(x^{2} + 1)}^{3} x + 4 y^{3} y' + \frac{d}{d x} [6 y^{2}] + \frac{d}{d x} [9]$

Step 2.7

Evaluate $\frac{d}{d x} [6 y^{2}]$ .

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Step 2.7.1

Since $6$ is constant with respect to $x$ , the derivative of $6 y^{2}$ with respect to $x$ is $6 \frac{d}{d x} [y^{2}]$ .

$16 {(x^{2} + 1)}^{3} x + 4 y^{3} y' + 6 \frac{d}{d x} [y^{2}] + \frac{d}{d x} [9]$

Step 2.7.2

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 2.7.2.1

To apply the Chain Rule, set $u_{3}$ as $y$ .

$16 {(x^{2} + 1)}^{3} x + 4 y^{3} y' + 6 (\frac{d}{d u_{3}} [{u_{3}}^{2}] \frac{d}{d x} [y]) + \frac{d}{d x} [9]$

Step 2.7.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{3}} [{u_{3}}^{n}]$ is $n {u_{3}}^{n - 1}$ where $n = 2$ .

$16 {(x^{2} + 1)}^{3} x + 4 y^{3} y' + 6 (2 u_{3} \frac{d}{d x} [y]) + \frac{d}{d x} [9]$

Step 2.7.2.3

Replace all occurrences of $u_{3}$ with $y$ .

$16 {(x^{2} + 1)}^{3} x + 4 y^{3} y' + 6 (2 y \frac{d}{d x} [y]) + \frac{d}{d x} [9]$

Step 2.7.3

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

$16 {(x^{2} + 1)}^{3} x + 4 y^{3} y' + 6 (2 y y') + \frac{d}{d x} [9]$

Step 2.7.4

Multiply $2$ by $6$ .

$16 {(x^{2} + 1)}^{3} x + 4 y^{3} y' + 12 y y' + \frac{d}{d x} [9]$

Step 2.8

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$16 {(x^{2} + 1)}^{3} x + 4 y^{3} y' + 12 y y' + 0$

Step 2.9

Simplify.

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Step 2.9.1

Add $16 {(x^{2} + 1)}^{3} x + 4 y^{3} y' + 12 y y'$ and $0$ .

$16 {(x^{2} + 1)}^{3} x + 4 y^{3} y' + 12 y y'$

Step 2.9.2

Reorder terms.

$16 x {(x^{2} + 1)}^{3} + 4 y^{3} y' + 12 y y'$

Step 3

Since $24$ is constant with respect to $x$ , the derivative of $24$ with respect to $x$ is $0$ .

$0$

Step 4

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

$16 x {(x^{2} + 1)}^{3} + 4 y^{3} y' + 12 y y' = 0$

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $16 x {(x^{2} + 1)}^{3}$ from both sides of the equation.

$4 y^{3} y' + 12 y y' = - 16 x {(x^{2} + 1)}^{3}$

Step 5.2

Factor $4 y y'$ out of $4 y^{3} y' + 12 y y'$ .

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Step 5.2.1

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

$4 y y' y^{2} + 12 y y' = - 16 x {(x^{2} + 1)}^{3}$

Step 5.2.2

Factor $4 y y'$ out of $12 y y'$ .

$4 y y' y^{2} + 4 y y' \cdot 3 = - 16 x {(x^{2} + 1)}^{3}$

Step 5.2.3

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

$4 y y' (y^{2} + 3) = - 16 x {(x^{2} + 1)}^{3}$

Step 5.3

Divide each term in $4 y y' (y^{2} + 3) = - 16 x {(x^{2} + 1)}^{3}$ by $4 y (y^{2} + 3)$ and simplify.

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Step 5.3.1

Divide each term in $4 y y' (y^{2} + 3) = - 16 x {(x^{2} + 1)}^{3}$ by $4 y (y^{2} + 3)$ .

$\frac{4 y y' (y^{2} + 3)}{4 y (y^{2} + 3)} = \frac{- 16 x {(x^{2} + 1)}^{3}}{4 y (y^{2} + 3)}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $4$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{4 y y' (y^{2} + 3)}{4 y (y^{2} + 3)} = \frac{- 16 x {(x^{2} + 1)}^{3}}{4 y (y^{2} + 3)}$

Step 5.3.2.1.2

Rewrite the expression.

$\frac{y y' (y^{2} + 3)}{y (y^{2} + 3)} = \frac{- 16 x {(x^{2} + 1)}^{3}}{4 y (y^{2} + 3)}$

Step 5.3.2.2

Cancel the common factor of $y$ .

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Step 5.3.2.2.1

Cancel the common factor.

$\frac{y y' (y^{2} + 3)}{y (y^{2} + 3)} = \frac{- 16 x {(x^{2} + 1)}^{3}}{4 y (y^{2} + 3)}$

Step 5.3.2.2.2

Rewrite the expression.

$\frac{y' (y^{2} + 3)}{y^{2} + 3} = \frac{- 16 x {(x^{2} + 1)}^{3}}{4 y (y^{2} + 3)}$

Step 5.3.2.3

Cancel the common factor of $y^{2} + 3$ .

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Step 5.3.2.3.1

Cancel the common factor.

$\frac{y' (y^{2} + 3)}{y^{2} + 3} = \frac{- 16 x {(x^{2} + 1)}^{3}}{4 y (y^{2} + 3)}$

Step 5.3.2.3.2

Divide $y'$ by $1$ .

$y' = \frac{- 16 x {(x^{2} + 1)}^{3}}{4 y (y^{2} + 3)}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Cancel the common factor of $- 16$ and $4$ .

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Step 5.3.3.1.1

Factor $4$ out of $- 16 x {(x^{2} + 1)}^{3}$ .

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

Step 5.3.3.1.2

Cancel the common factors.

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Step 5.3.3.1.2.1

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

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

Step 5.3.3.1.2.2

Cancel the common factor.

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

Step 5.3.3.1.2.3

Rewrite the expression.

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

Step 5.3.3.2

Move the negative in front of the fraction.

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

Step 6

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

$\frac{d y}{d x} = - \frac{4 x {(x^{2} + 1)}^{3}}{y (y^{2} + 3)}$