Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

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

$1$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Simplify each term.

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Step 3.1.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 3.1.2

Combine $2$ and $\frac{1}{x^{2}}$ .

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

Step 3.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 3.3

Simplify the denominator.

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Step 3.3.1

To write $3$ as a fraction with a common denominator, multiply by $\frac{x^{2}}{x^{2}}$ .

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

Step 3.3.2

Combine the numerators over the common denominator.

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

Step 3.3.3

Apply the product rule to $\frac{2 + 3 x^{2}}{x^{2}}$ .

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

Step 3.3.4

Multiply the exponents in ${(x^{2})}^{3}$ .

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Step 3.3.4.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.3.4.2

Multiply $2$ by $3$ .

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

Step 3.4

Multiply the numerator by the reciprocal of the denominator.

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

Step 3.5

Multiply $\frac{x^{6}}{{(2 + 3 x^{2})}^{3}}$ by $1$ .

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

Step 3.6

Differentiate using the Quotient Rule which states that $\frac{d}{d y} [\frac{f (y)}{g (y)}]$ is $\frac{g (y) \frac{d}{d y} [f (y)] - f (y) \frac{d}{d y} [g (y)]}{{g (y)}^{2}}$ where $f (y) = x^{6}$ and $g (y) = {(2 + 3 x^{2})}^{3}$ .

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

Step 3.7

Multiply the exponents in ${({(2 + 3 x^{2})}^{3})}^{2}$ .

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Step 3.7.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{{(2 + 3 x^{2})}^{3} \frac{d}{d y} [x^{6}] - x^{6} \frac{d}{d y} [{(2 + 3 x^{2})}^{3}]}{{(2 + 3 x^{2})}^{3 \cdot 2}}$

Step 3.7.2

Multiply $3$ by $2$ .

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

Step 3.8

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{6}$ and $g (y) = x$ .

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Step 3.8.1

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

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

Step 3.8.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 6$ .

$\frac{{(2 + 3 x^{2})}^{3} (6 {u_{1}}^{5} \frac{d}{d y} [x]) - x^{6} \frac{d}{d y} [{(2 + 3 x^{2})}^{3}]}{{(2 + 3 x^{2})}^{6}}$

Step 3.8.3

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

$\frac{{(2 + 3 x^{2})}^{3} (6 x^{5} \frac{d}{d y} [x]) - x^{6} \frac{d}{d y} [{(2 + 3 x^{2})}^{3}]}{{(2 + 3 x^{2})}^{6}}$

Step 3.9

Move $6$ to the left of ${(2 + 3 x^{2})}^{3}$ .

$\frac{6 \cdot {(2 + 3 x^{2})}^{3} x^{5} \frac{d}{d y} [x] - x^{6} \frac{d}{d y} [{(2 + 3 x^{2})}^{3}]}{{(2 + 3 x^{2})}^{6}}$

Step 3.10

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

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - x^{6} \frac{d}{d y} [{(2 + 3 x^{2})}^{3}]}{{(2 + 3 x^{2})}^{6}}$

Step 3.11

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = y^{3}$ and $g (y) = 2 + 3 x^{2}$ .

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Step 3.11.1

To apply the Chain Rule, set $u_{2}$ as $2 + 3 x^{2}$ .

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - x^{6} (\frac{d}{d u_{2}} [{u_{2}}^{3}] \frac{d}{d y} [2 + 3 x^{2}])}{{(2 + 3 x^{2})}^{6}}$

Step 3.11.2

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

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - x^{6} (3 {u_{2}}^{2} \frac{d}{d y} [2 + 3 x^{2}])}{{(2 + 3 x^{2})}^{6}}$

Step 3.11.3

Replace all occurrences of $u_{2}$ with $2 + 3 x^{2}$ .

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - x^{6} (3 {(2 + 3 x^{2})}^{2} \frac{d}{d y} [2 + 3 x^{2}])}{{(2 + 3 x^{2})}^{6}}$

Step 3.12

Differentiate.

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Step 3.12.1

Multiply $3$ by $- 1$ .

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 3 x^{6} ({(2 + 3 x^{2})}^{2} \frac{d}{d y} [2 + 3 x^{2}])}{{(2 + 3 x^{2})}^{6}}$

Step 3.12.2

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

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 3 x^{6} ({(2 + 3 x^{2})}^{2} (\frac{d}{d y} [2] + \frac{d}{d y} [3 x^{2}]))}{{(2 + 3 x^{2})}^{6}}$

Step 3.12.3

Since $2$ is constant with respect to $y$ , the derivative of $2$ with respect to $y$ is $0$ .

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 3 x^{6} ({(2 + 3 x^{2})}^{2} (0 + \frac{d}{d y} [3 x^{2}]))}{{(2 + 3 x^{2})}^{6}}$

Step 3.12.4

Add $0$ and $\frac{d}{d y} [3 x^{2}]$ .

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 3 x^{6} ({(2 + 3 x^{2})}^{2} \frac{d}{d y} [3 x^{2}])}{{(2 + 3 x^{2})}^{6}}$

Step 3.12.5

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

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 3 x^{6} ({(2 + 3 x^{2})}^{2} (3 \frac{d}{d y} [x^{2}]))}{{(2 + 3 x^{2})}^{6}}$

Step 3.12.6

Simplify the expression.

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Step 3.12.6.1

Move $3$ to the left of ${(2 + 3 x^{2})}^{2}$ .

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 3 x^{6} (3 \cdot {(2 + 3 x^{2})}^{2} \frac{d}{d y} [x^{2}])}{{(2 + 3 x^{2})}^{6}}$

Step 3.12.6.2

Multiply $3$ by $- 3$ .

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 9 x^{6} ({(2 + 3 x^{2})}^{2} \frac{d}{d y} [x^{2}])}{{(2 + 3 x^{2})}^{6}}$

Step 3.13

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

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Step 3.13.1

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

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 9 x^{6} ({(2 + 3 x^{2})}^{2} (\frac{d}{d u_{3}} [{u_{3}}^{2}] \frac{d}{d y} [x]))}{{(2 + 3 x^{2})}^{6}}$

Step 3.13.2

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

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 9 x^{6} ({(2 + 3 x^{2})}^{2} (2 u_{3} \frac{d}{d y} [x]))}{{(2 + 3 x^{2})}^{6}}$

Step 3.13.3

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

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 9 x^{6} ({(2 + 3 x^{2})}^{2} (2 x \frac{d}{d y} [x]))}{{(2 + 3 x^{2})}^{6}}$

Step 3.14

Simplify the expression.

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Step 3.14.1

Move $2$ to the left of ${(2 + 3 x^{2})}^{2}$ .

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 9 x^{6} (2 \cdot {(2 + 3 x^{2})}^{2} x \frac{d}{d y} [x])}{{(2 + 3 x^{2})}^{6}}$

Step 3.14.2

Multiply $2$ by $- 9$ .

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 18 x^{6} ({(2 + 3 x^{2})}^{2} x \frac{d}{d y} [x])}{{(2 + 3 x^{2})}^{6}}$

Step 3.15

Raise $x$ to the power of $1$ .

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 18 (x^{1} x^{6}) ({(2 + 3 x^{2})}^{2} \frac{d}{d y} [x])}{{(2 + 3 x^{2})}^{6}}$

Step 3.16

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

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 18 x^{1 + 6} ({(2 + 3 x^{2})}^{2} \frac{d}{d y} [x])}{{(2 + 3 x^{2})}^{6}}$

Step 3.17

Add $1$ and $6$ .

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 18 x^{7} ({(2 + 3 x^{2})}^{2} \frac{d}{d y} [x])}{{(2 + 3 x^{2})}^{6}}$

Step 3.18

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

$\frac{6 {(2 + 3 x^{2})}^{3} x^{5} x' - 18 x^{7} ({(2 + 3 x^{2})}^{2} x')}{{(2 + 3 x^{2})}^{6}}$

Step 3.19

Simplify.

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Step 3.19.1

Simplify the numerator.

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Step 3.19.1.1

Factor $6 {(2 + 3 x^{2})}^{2} x^{5} x'$ out of $6 {(2 + 3 x^{2})}^{3} x^{5} x' - 18 x^{7} ({(2 + 3 x^{2})}^{2} x')$ .

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Step 3.19.1.1.1

Factor $6 {(2 + 3 x^{2})}^{2} x^{5} x'$ out of $6 {(2 + 3 x^{2})}^{3} x^{5} x'$ .

$\frac{6 {(2 + 3 x^{2})}^{2} x^{5} x' (2 + 3 x^{2}) - 18 x^{7} ({(2 + 3 x^{2})}^{2} x')}{{(2 + 3 x^{2})}^{6}}$

Step 3.19.1.1.2

Factor $6 {(2 + 3 x^{2})}^{2} x^{5} x'$ out of $- 18 x^{7} ({(2 + 3 x^{2})}^{2} x')$ .

$\frac{6 {(2 + 3 x^{2})}^{2} x^{5} x' (2 + 3 x^{2}) + 6 {(2 + 3 x^{2})}^{2} x^{5} x' (- 3 x^{2})}{{(2 + 3 x^{2})}^{6}}$

Step 3.19.1.1.3

Factor $6 {(2 + 3 x^{2})}^{2} x^{5} x'$ out of $6 {(2 + 3 x^{2})}^{2} x^{5} x' (2 + 3 x^{2}) + 6 {(2 + 3 x^{2})}^{2} x^{5} x' (- 3 x^{2})$ .

$\frac{6 {(2 + 3 x^{2})}^{2} x^{5} x' (2 + 3 x^{2} - 3 x^{2})}{{(2 + 3 x^{2})}^{6}}$

Step 3.19.1.2

Subtract $3 x^{2}$ from $3 x^{2}$ .

$\frac{6 {(2 + 3 x^{2})}^{2} x^{5} x' (2 + 0)}{{(2 + 3 x^{2})}^{6}}$

Step 3.19.1.3

Add $2$ and $0$ .

$\frac{6 {(2 + 3 x^{2})}^{2} x^{5} x' \cdot 2}{{(2 + 3 x^{2})}^{6}}$

Step 3.19.1.4

Multiply $2$ by $6$ .

$\frac{12 {(2 + 3 x^{2})}^{2} x^{5} x'}{{(2 + 3 x^{2})}^{6}}$

Step 3.19.2

Cancel the common factor of ${(2 + 3 x^{2})}^{2}$ and ${(2 + 3 x^{2})}^{6}$ .

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Step 3.19.2.1

Factor ${(2 + 3 x^{2})}^{2}$ out of $12 {(2 + 3 x^{2})}^{2} x^{5} x'$ .

$\frac{{(2 + 3 x^{2})}^{2} (12 x^{5} x')}{{(2 + 3 x^{2})}^{6}}$

Step 3.19.2.2

Cancel the common factors.

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Step 3.19.2.2.1

Factor ${(2 + 3 x^{2})}^{2}$ out of ${(2 + 3 x^{2})}^{6}$ .

$\frac{{(2 + 3 x^{2})}^{2} (12 x^{5} x')}{{(2 + 3 x^{2})}^{2} {(2 + 3 x^{2})}^{4}}$

Step 3.19.2.2.2

Cancel the common factor.

$\frac{{(2 + 3 x^{2})}^{2} (12 x^{5} x')}{{(2 + 3 x^{2})}^{2} {(2 + 3 x^{2})}^{4}}$

Step 3.19.2.2.3

Rewrite the expression.

$\frac{12 x^{5} x'}{{(2 + 3 x^{2})}^{4}}$

Step 3.19.3

Reorder terms.

$\frac{12 x^{5} x'}{{(3 x^{2} + 2)}^{4}}$

Step 4

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

$1 = \frac{12 x^{5} x'}{{(3 x^{2} + 2)}^{4}}$

Step 5

Solve for $x'$ .

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Step 5.1

Rewrite the equation as $\frac{12 x^{5} x'}{{(3 x^{2} + 2)}^{4}} = 1$ .

$\frac{12 x^{5} x'}{{(3 x^{2} + 2)}^{4}} = 1$

Step 5.2

Multiply both sides by ${(3 x^{2} + 2)}^{4}$ .

$\frac{12 x^{5} x'}{{(3 x^{2} + 2)}^{4}} {(3 x^{2} + 2)}^{4} = 1 {(3 x^{2} + 2)}^{4}$

Step 5.3

Simplify.

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Step 5.3.1

Simplify the left side.

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Step 5.3.1.1

Cancel the common factor of ${(3 x^{2} + 2)}^{4}$ .

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Step 5.3.1.1.1

Cancel the common factor.

$\frac{12 x^{5} x'}{{(3 x^{2} + 2)}^{4}} {(3 x^{2} + 2)}^{4} = 1 {(3 x^{2} + 2)}^{4}$

Step 5.3.1.1.2

Rewrite the expression.

$12 x^{5} x' = 1 {(3 x^{2} + 2)}^{4}$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Multiply ${(3 x^{2} + 2)}^{4}$ by $1$ .

$12 x^{5} x' = {(3 x^{2} + 2)}^{4}$

Step 5.4

Divide each term in $12 x^{5} x' = {(3 x^{2} + 2)}^{4}$ by $12 x^{5}$ and simplify.

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Step 5.4.1

Divide each term in $12 x^{5} x' = {(3 x^{2} + 2)}^{4}$ by $12 x^{5}$ .

$\frac{12 x^{5} x'}{12 x^{5}} = \frac{{(3 x^{2} + 2)}^{4}}{12 x^{5}}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $12$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{12 x^{5} x'}{12 x^{5}} = \frac{{(3 x^{2} + 2)}^{4}}{12 x^{5}}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{x^{5} x'}{x^{5}} = \frac{{(3 x^{2} + 2)}^{4}}{12 x^{5}}$

Step 5.4.2.2

Cancel the common factor of $x^{5}$ .

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Step 5.4.2.2.1

Cancel the common factor.

$\frac{x^{5} x'}{x^{5}} = \frac{{(3 x^{2} + 2)}^{4}}{12 x^{5}}$

Step 5.4.2.2.2

Divide $x'$ by $1$ .

$x' = \frac{{(3 x^{2} + 2)}^{4}}{12 x^{5}}$

Step 6

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

$\frac{d x}{d y} = \frac{{(3 x^{2} + 2)}^{4}}{12 x^{5}}$