Calculus Examples

$\frac{{(5 t - 2)}^{2}}{{(t^{2} + 1)}^{3}}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d t} [\frac{f (t)}{g (t)}]$ is $\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}$ where $f (t) = {(5 t - 2)}^{2}$ and $g (t) = {(t^{2} + 1)}^{3}$ .

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

Step 2

Multiply the exponents in ${({(t^{2} + 1)}^{3})}^{2}$ .

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Step 2.1

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

$\frac{{(t^{2} + 1)}^{3} \frac{d}{d t} [{(5 t - 2)}^{2}] - {(5 t - 2)}^{2} \frac{d}{d t} [{(t^{2} + 1)}^{3}]}{{(t^{2} + 1)}^{3 \cdot 2}}$

Step 2.2

Multiply $3$ by $2$ .

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

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{2}$ and $g (t) = 5 t - 2$ .

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Step 3.1

To apply the Chain Rule, set $u_{1}$ as $5 t - 2$ .

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

Step 3.2

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

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

Step 3.3

Replace all occurrences of $u_{1}$ with $5 t - 2$ .

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

Step 4

Differentiate.

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Step 4.1

Move $2$ to the left of ${(t^{2} + 1)}^{3}$ .

$\frac{2 \cdot {(t^{2} + 1)}^{3} (5 t - 2) \frac{d}{d t} [5 t - 2] - {(5 t - 2)}^{2} \frac{d}{d t} [{(t^{2} + 1)}^{3}]}{{(t^{2} + 1)}^{6}}$

Step 4.2

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

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

Step 4.3

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

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

Step 4.4

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

$\frac{2 {(t^{2} + 1)}^{3} (5 t - 2) (5 \cdot 1 + \frac{d}{d t} [- 2]) - {(5 t - 2)}^{2} \frac{d}{d t} [{(t^{2} + 1)}^{3}]}{{(t^{2} + 1)}^{6}}$

Step 4.5

Multiply $5$ by $1$ .

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

Step 4.6

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

$\frac{2 {(t^{2} + 1)}^{3} (5 t - 2) (5 + 0) - {(5 t - 2)}^{2} \frac{d}{d t} [{(t^{2} + 1)}^{3}]}{{(t^{2} + 1)}^{6}}$

Step 4.7

Simplify the expression.

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Step 4.7.1

Add $5$ and $0$ .

$\frac{2 {(t^{2} + 1)}^{3} (5 t - 2) \cdot 5 - {(5 t - 2)}^{2} \frac{d}{d t} [{(t^{2} + 1)}^{3}]}{{(t^{2} + 1)}^{6}}$

Step 4.7.2

Multiply $5$ by $2$ .

$\frac{10 {(t^{2} + 1)}^{3} (5 t - 2) - {(5 t - 2)}^{2} \frac{d}{d t} [{(t^{2} + 1)}^{3}]}{{(t^{2} + 1)}^{6}}$

Step 5

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{3}$ and $g (t) = t^{2} + 1$ .

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Step 5.1

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

$\frac{10 {(t^{2} + 1)}^{3} (5 t - 2) - {(5 t - 2)}^{2} (\frac{d}{d u_{2}} [{u_{2}}^{3}] \frac{d}{d t} [t^{2} + 1])}{{(t^{2} + 1)}^{6}}$

Step 5.2

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

$\frac{10 {(t^{2} + 1)}^{3} (5 t - 2) - {(5 t - 2)}^{2} (3 {u_{2}}^{2} \frac{d}{d t} [t^{2} + 1])}{{(t^{2} + 1)}^{6}}$

Step 5.3

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

$\frac{10 {(t^{2} + 1)}^{3} (5 t - 2) - {(5 t - 2)}^{2} (3 {(t^{2} + 1)}^{2} \frac{d}{d t} [t^{2} + 1])}{{(t^{2} + 1)}^{6}}$

Step 6

Differentiate.

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Step 6.1

Multiply $3$ by $- 1$ .

$\frac{10 {(t^{2} + 1)}^{3} (5 t - 2) - 3 {(5 t - 2)}^{2} ({(t^{2} + 1)}^{2} \frac{d}{d t} [t^{2} + 1])}{{(t^{2} + 1)}^{6}}$

Step 6.2

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

$\frac{10 {(t^{2} + 1)}^{3} (5 t - 2) - 3 {(5 t - 2)}^{2} ({(t^{2} + 1)}^{2} (\frac{d}{d t} [t^{2}] + \frac{d}{d t} [1]))}{{(t^{2} + 1)}^{6}}$

Step 6.3

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

$\frac{10 {(t^{2} + 1)}^{3} (5 t - 2) - 3 {(5 t - 2)}^{2} ({(t^{2} + 1)}^{2} (2 t + \frac{d}{d t} [1]))}{{(t^{2} + 1)}^{6}}$

Step 6.4

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

$\frac{10 {(t^{2} + 1)}^{3} (5 t - 2) - 3 {(5 t - 2)}^{2} ({(t^{2} + 1)}^{2} (2 t + 0))}{{(t^{2} + 1)}^{6}}$

Step 6.5

Simplify the expression.

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Step 6.5.1

Add $2 t$ and $0$ .

$\frac{10 {(t^{2} + 1)}^{3} (5 t - 2) - 3 {(5 t - 2)}^{2} ({(t^{2} + 1)}^{2} (2 t))}{{(t^{2} + 1)}^{6}}$

Step 6.5.2

Move $2$ to the left of ${(t^{2} + 1)}^{2}$ .

$\frac{10 {(t^{2} + 1)}^{3} (5 t - 2) - 3 {(5 t - 2)}^{2} (2 \cdot {(t^{2} + 1)}^{2} t)}{{(t^{2} + 1)}^{6}}$

Step 6.5.3

Multiply $2$ by $- 3$ .

$\frac{10 {(t^{2} + 1)}^{3} (5 t - 2) - 6 {(5 t - 2)}^{2} ({(t^{2} + 1)}^{2} t)}{{(t^{2} + 1)}^{6}}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

Factor $2 {(t^{2} + 1)}^{2} (5 t - 2)$ out of $10 {(t^{2} + 1)}^{3} (5 t - 2) - 6 {(5 t - 2)}^{2} ({(t^{2} + 1)}^{2} t)$ .

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Step 7.1.1.1

Factor $2 {(t^{2} + 1)}^{2} (5 t - 2)$ out of $10 {(t^{2} + 1)}^{3} (5 t - 2)$ .

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (5 (t^{2} + 1)) - 6 {(5 t - 2)}^{2} ({(t^{2} + 1)}^{2} t)}{{(t^{2} + 1)}^{6}}$

Step 7.1.1.2

Factor $2 {(t^{2} + 1)}^{2} (5 t - 2)$ out of $- 6 {(5 t - 2)}^{2} ({(t^{2} + 1)}^{2} t)$ .

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (5 (t^{2} + 1)) + 2 {(t^{2} + 1)}^{2} (5 t - 2) (- 3 (5 t - 2) t)}{{(t^{2} + 1)}^{6}}$

Step 7.1.1.3

Factor $2 {(t^{2} + 1)}^{2} (5 t - 2)$ out of $2 {(t^{2} + 1)}^{2} (5 t - 2) (5 (t^{2} + 1)) + 2 {(t^{2} + 1)}^{2} (5 t - 2) (- 3 (5 t - 2) t)$ .

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (5 (t^{2} + 1) - 3 (5 t - 2) t)}{{(t^{2} + 1)}^{6}}$

Step 7.1.2

Apply the distributive property.

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (5 t^{2} + 5 \cdot 1 - 3 (5 t - 2) t)}{{(t^{2} + 1)}^{6}}$

Step 7.1.3

Multiply $5$ by $1$ .

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (5 t^{2} + 5 - 3 (5 t - 2) t)}{{(t^{2} + 1)}^{6}}$

Step 7.1.4

Apply the distributive property.

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (5 t^{2} + 5 + (- 3 (5 t) - 3 \cdot - 2) t)}{{(t^{2} + 1)}^{6}}$

Step 7.1.5

Multiply $5$ by $- 3$ .

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (5 t^{2} + 5 + (- 15 t - 3 \cdot - 2) t)}{{(t^{2} + 1)}^{6}}$

Step 7.1.6

Multiply $- 3$ by $- 2$ .

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (5 t^{2} + 5 + (- 15 t + 6) t)}{{(t^{2} + 1)}^{6}}$

Step 7.1.7

Apply the distributive property.

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (5 t^{2} + 5 - 15 t \cdot t + 6 t)}{{(t^{2} + 1)}^{6}}$

Step 7.1.8

Multiply $t$ by $t$ by adding the exponents.

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Step 7.1.8.1

Move $t$ .

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (5 t^{2} + 5 - 15 (t \cdot t) + 6 t)}{{(t^{2} + 1)}^{6}}$

Step 7.1.8.2

Multiply $t$ by $t$ .

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (5 t^{2} + 5 - 15 t^{2} + 6 t)}{{(t^{2} + 1)}^{6}}$

Step 7.1.9

Subtract $15 t^{2}$ from $5 t^{2}$ .

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (- 10 t^{2} + 5 + 6 t)}{{(t^{2} + 1)}^{6}}$

Step 7.1.10

Reorder terms.

$\frac{2 {(t^{2} + 1)}^{2} (5 t - 2) (- 10 t^{2} + 6 t + 5)}{{(t^{2} + 1)}^{6}}$

Step 7.2

Cancel the common factor of ${(t^{2} + 1)}^{2}$ and ${(t^{2} + 1)}^{6}$ .

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Step 7.2.1

Factor ${(t^{2} + 1)}^{2}$ out of $2 {(t^{2} + 1)}^{2} (5 t - 2) (- 10 t^{2} + 6 t + 5)$ .

$\frac{{(t^{2} + 1)}^{2} ((2 (5 t - 2)) (- 10 t^{2} + 6 t + 5))}{{(t^{2} + 1)}^{6}}$

Step 7.2.2

Cancel the common factors.

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Step 7.2.2.1

Factor ${(t^{2} + 1)}^{2}$ out of ${(t^{2} + 1)}^{6}$ .

$\frac{{(t^{2} + 1)}^{2} ((2 (5 t - 2)) (- 10 t^{2} + 6 t + 5))}{{(t^{2} + 1)}^{2} {(t^{2} + 1)}^{4}}$

Step 7.2.2.2

Cancel the common factor.

$\frac{{(t^{2} + 1)}^{2} ((2 (5 t - 2)) (- 10 t^{2} + 6 t + 5))}{{(t^{2} + 1)}^{2} {(t^{2} + 1)}^{4}}$

Step 7.2.2.3

Rewrite the expression.

$\frac{(2 (5 t - 2)) (- 10 t^{2} + 6 t + 5)}{{(t^{2} + 1)}^{4}}$

$\frac{2 (5 t - 2) (- 10 t^{2} + 6 t + 5)}{{(t^{2} + 1)}^{4}}$

Step 7.3

Reorder terms.

$\frac{2 (- 10 t^{2} + 6 t + 5) (5 t - 2)}{{(t^{2} + 1)}^{4}}$

Step 7.4

Factor $- 1$ out of $- 10 t^{2}$ .

$\frac{2 (- (10 t^{2}) + 6 t + 5) (5 t - 2)}{{(t^{2} + 1)}^{4}}$

Step 7.5

Factor $- 1$ out of $6 t$ .

$\frac{2 (- (10 t^{2}) - (- 6 t) + 5) (5 t - 2)}{{(t^{2} + 1)}^{4}}$

Step 7.6

Factor $- 1$ out of $- (10 t^{2}) - (- 6 t)$ .

$\frac{2 (- (10 t^{2} - 6 t) + 5) (5 t - 2)}{{(t^{2} + 1)}^{4}}$

Step 7.7

Rewrite $5$ as $- 1 (- 5)$ .

$\frac{2 (- (10 t^{2} - 6 t) - 1 (- 5)) (5 t - 2)}{{(t^{2} + 1)}^{4}}$

Step 7.8

Factor $- 1$ out of $- (10 t^{2} - 6 t) - 1 (- 5)$ .

$\frac{2 (- (10 t^{2} - 6 t - 5)) (5 t - 2)}{{(t^{2} + 1)}^{4}}$

Step 7.9

Rewrite $- (10 t^{2} - 6 t - 5)$ as $- 1 (10 t^{2} - 6 t - 5)$ .

$\frac{2 (- 1 (10 t^{2} - 6 t - 5)) (5 t - 2)}{{(t^{2} + 1)}^{4}}$

Step 7.10

Move the negative in front of the fraction.

$- \frac{(2 (10 t^{2} - 6 t - 5)) (5 t - 2)}{{(t^{2} + 1)}^{4}}$

Step 7.11

Reorder factors in $- \frac{(2 (10 t^{2} - 6 t - 5)) (5 t - 2)}{{(t^{2} + 1)}^{4}}$ .

$- \frac{2 (10 t^{2} - 6 t - 5) (5 t - 2)}{{(t^{2} + 1)}^{4}}$