Calculus Examples

$y = \frac{t^{2} + 3}{{(5 t - 2)}^{9}}$

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 $f (t) = t^{2} + 3$ and $g (t) = {(5 t - 2)}^{9}$ .

$\frac{{(5 t - 2)}^{9} \frac{d}{d t} [t^{2} + 3] - (t^{2} + 3) \frac{d}{d t} [{(5 t - 2)}^{9}]}{{({(5 t - 2)}^{9})}^{2}}$

Step 2

Differentiate.

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Step 2.1

Multiply the exponents in ${({(5 t - 2)}^{9})}^{2}$ .

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Step 2.1.1

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

$\frac{{(5 t - 2)}^{9} \frac{d}{d t} [t^{2} + 3] - (t^{2} + 3) \frac{d}{d t} [{(5 t - 2)}^{9}]}{{(5 t - 2)}^{9 \cdot 2}}$

Step 2.1.2

Multiply $9$ by $2$ .

$\frac{{(5 t - 2)}^{9} \frac{d}{d t} [t^{2} + 3] - (t^{2} + 3) \frac{d}{d t} [{(5 t - 2)}^{9}]}{{(5 t - 2)}^{18}}$

Step 2.2

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

$\frac{{(5 t - 2)}^{9} (\frac{d}{d t} [t^{2}] + \frac{d}{d t} [3]) - (t^{2} + 3) \frac{d}{d t} [{(5 t - 2)}^{9}]}{{(5 t - 2)}^{18}}$

Step 2.3

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

$\frac{{(5 t - 2)}^{9} (2 t + \frac{d}{d t} [3]) - (t^{2} + 3) \frac{d}{d t} [{(5 t - 2)}^{9}]}{{(5 t - 2)}^{18}}$

Step 2.4

Since $3$ is constant with respect to $t$ , the derivative of $3$ with respect to $t$ is $0$ .

$\frac{{(5 t - 2)}^{9} (2 t + 0) - (t^{2} + 3) \frac{d}{d t} [{(5 t - 2)}^{9}]}{{(5 t - 2)}^{18}}$

Step 2.5

Simplify the expression.

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Step 2.5.1

Add $2 t$ and $0$ .

$\frac{{(5 t - 2)}^{9} (2 t) - (t^{2} + 3) \frac{d}{d t} [{(5 t - 2)}^{9}]}{{(5 t - 2)}^{18}}$

Step 2.5.2

Move $2$ to the left of ${(5 t - 2)}^{9}$ .

$\frac{2 \cdot {(5 t - 2)}^{9} t - (t^{2} + 3) \frac{d}{d t} [{(5 t - 2)}^{9}]}{{(5 t - 2)}^{18}}$

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{9}$ and $g (t) = 5 t - 2$ .

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Step 3.1

To apply the Chain Rule, set $u$ as $5 t - 2$ .

$\frac{2 {(5 t - 2)}^{9} t - (t^{2} + 3) (\frac{d}{d u} [u^{9}] \frac{d}{d t} [5 t - 2])}{{(5 t - 2)}^{18}}$

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 9$ .

$\frac{2 {(5 t - 2)}^{9} t - (t^{2} + 3) (9 u^{8} \frac{d}{d t} [5 t - 2])}{{(5 t - 2)}^{18}}$

Step 3.3

Replace all occurrences of $u$ with $5 t - 2$ .

$\frac{2 {(5 t - 2)}^{9} t - (t^{2} + 3) (9 {(5 t - 2)}^{8} \frac{d}{d t} [5 t - 2])}{{(5 t - 2)}^{18}}$

Step 4

Simplify with factoring out.

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Step 4.1

Multiply $9$ by $- 1$ .

$\frac{2 {(5 t - 2)}^{9} t - 9 (t^{2} + 3) ({(5 t - 2)}^{8} \frac{d}{d t} [5 t - 2])}{{(5 t - 2)}^{18}}$

Step 4.2

Factor ${(5 t - 2)}^{8}$ out of $2 {(5 t - 2)}^{9} t - 9 (t^{2} + 3) ({(5 t - 2)}^{8} \frac{d}{d t} [5 t - 2])$ .

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Step 4.2.1

Factor ${(5 t - 2)}^{8}$ out of $2 {(5 t - 2)}^{9} t$ .

$\frac{{(5 t - 2)}^{8} (2 (5 t - 2) t) - 9 (t^{2} + 3) ({(5 t - 2)}^{8} \frac{d}{d t} [5 t - 2])}{{(5 t - 2)}^{18}}$

Step 4.2.2

Factor ${(5 t - 2)}^{8}$ out of $- 9 (t^{2} + 3) ({(5 t - 2)}^{8} \frac{d}{d t} [5 t - 2])$ .

$\frac{{(5 t - 2)}^{8} (2 (5 t - 2) t) + {(5 t - 2)}^{8} (- 9 (t^{2} + 3) (\frac{d}{d t} [5 t - 2]))}{{(5 t - 2)}^{18}}$

Step 4.2.3

Factor ${(5 t - 2)}^{8}$ out of ${(5 t - 2)}^{8} (2 (5 t - 2) t) + {(5 t - 2)}^{8} (- 9 (t^{2} + 3) (\frac{d}{d t} [5 t - 2]))$ .

$\frac{{(5 t - 2)}^{8} (2 (5 t - 2) t - 9 (t^{2} + 3) (\frac{d}{d t} [5 t - 2]))}{{(5 t - 2)}^{18}}$

Step 5

Cancel the common factors.

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Step 5.1

Factor ${(5 t - 2)}^{8}$ out of ${(5 t - 2)}^{18}$ .

$\frac{{(5 t - 2)}^{8} (2 (5 t - 2) t - 9 (t^{2} + 3) (\frac{d}{d t} [5 t - 2]))}{{(5 t - 2)}^{8} {(5 t - 2)}^{10}}$

Step 5.2

Cancel the common factor.

$\frac{{(5 t - 2)}^{8} (2 (5 t - 2) t - 9 (t^{2} + 3) (\frac{d}{d t} [5 t - 2]))}{{(5 t - 2)}^{8} {(5 t - 2)}^{10}}$

Step 5.3

Rewrite the expression.

$\frac{2 (5 t - 2) t - 9 (t^{2} + 3) (\frac{d}{d t} [5 t - 2])}{{(5 t - 2)}^{10}}$

Step 6

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

$\frac{2 (5 t - 2) t - 9 (t^{2} + 3) (\frac{d}{d t} [5 t] + \frac{d}{d t} [- 2])}{{(5 t - 2)}^{10}}$

Step 7

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

$\frac{2 (5 t - 2) t - 9 (t^{2} + 3) (5 \frac{d}{d t} [t] + \frac{d}{d t} [- 2])}{{(5 t - 2)}^{10}}$

Step 8

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

$\frac{2 (5 t - 2) t - 9 (t^{2} + 3) (5 \cdot 1 + \frac{d}{d t} [- 2])}{{(5 t - 2)}^{10}}$

Step 9

Multiply $5$ by $1$ .

$\frac{2 (5 t - 2) t - 9 (t^{2} + 3) (5 + \frac{d}{d t} [- 2])}{{(5 t - 2)}^{10}}$

Step 10

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

$\frac{2 (5 t - 2) t - 9 (t^{2} + 3) (5 + 0)}{{(5 t - 2)}^{10}}$

Step 11

Simplify the expression.

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Step 11.1

Add $5$ and $0$ .

$\frac{2 (5 t - 2) t - 9 (t^{2} + 3) \cdot 5}{{(5 t - 2)}^{10}}$

Step 11.2

Multiply $5$ by $- 9$ .

$\frac{2 (5 t - 2) t - 45 (t^{2} + 3)}{{(5 t - 2)}^{10}}$

Step 12

Simplify.

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Step 12.1

Apply the distributive property.

$\frac{(2 (5 t) + 2 \cdot - 2) t - 45 (t^{2} + 3)}{{(5 t - 2)}^{10}}$

Step 12.2

Apply the distributive property.

$\frac{2 (5 t) t + 2 \cdot - 2 t - 45 (t^{2} + 3)}{{(5 t - 2)}^{10}}$

Step 12.3

Apply the distributive property.

$\frac{2 (5 t) t + 2 \cdot - 2 t - 45 t^{2} - 45 \cdot 3}{{(5 t - 2)}^{10}}$

Step 12.4

Simplify the numerator.

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Step 12.4.1

Simplify each term.

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Step 12.4.1.1

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

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Step 12.4.1.1.1

Move $t$ .

$\frac{2 (5 (t \cdot t)) + 2 \cdot - 2 t - 45 t^{2} - 45 \cdot 3}{{(5 t - 2)}^{10}}$

Step 12.4.1.1.2

Multiply $t$ by $t$ .

$\frac{2 (5 t^{2}) + 2 \cdot - 2 t - 45 t^{2} - 45 \cdot 3}{{(5 t - 2)}^{10}}$

Step 12.4.1.2

Multiply $5$ by $2$ .

$\frac{10 t^{2} + 2 \cdot - 2 t - 45 t^{2} - 45 \cdot 3}{{(5 t - 2)}^{10}}$

Step 12.4.1.3

Multiply $2$ by $- 2$ .

$\frac{10 t^{2} - 4 t - 45 t^{2} - 45 \cdot 3}{{(5 t - 2)}^{10}}$

Step 12.4.1.4

Multiply $- 45$ by $3$ .

$\frac{10 t^{2} - 4 t - 45 t^{2} - 135}{{(5 t - 2)}^{10}}$

Step 12.4.2

Subtract $45 t^{2}$ from $10 t^{2}$ .

$\frac{- 35 t^{2} - 4 t - 135}{{(5 t - 2)}^{10}}$

Step 12.5

Factor $- 1$ out of $- 35 t^{2}$ .

$\frac{- (35 t^{2}) - 4 t - 135}{{(5 t - 2)}^{10}}$

Step 12.6

Factor $- 1$ out of $- 4 t$ .

$\frac{- (35 t^{2}) - (4 t) - 135}{{(5 t - 2)}^{10}}$

Step 12.7

Factor $- 1$ out of $- (35 t^{2}) - (4 t)$ .

$\frac{- (35 t^{2} + 4 t) - 135}{{(5 t - 2)}^{10}}$

Step 12.8

Rewrite $- 135$ as $- 1 (135)$ .

$\frac{- (35 t^{2} + 4 t) - 1 (135)}{{(5 t - 2)}^{10}}$

Step 12.9

Factor $- 1$ out of $- (35 t^{2} + 4 t) - 1 (135)$ .

$\frac{- (35 t^{2} + 4 t + 135)}{{(5 t - 2)}^{10}}$

Step 12.10

Rewrite $- (35 t^{2} + 4 t + 135)$ as $- 1 (35 t^{2} + 4 t + 135)$ .

$\frac{- 1 (35 t^{2} + 4 t + 135)}{{(5 t - 2)}^{10}}$

Step 12.11

Move the negative in front of the fraction.

$- \frac{35 t^{2} + 4 t + 135}{{(5 t - 2)}^{10}}$