Calculus Examples

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

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

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

Step 2

Differentiate.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

Multiply $9$ by $2$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Simplify the expression.

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Step 2.5.1

Add $2 t$ and $0$ .

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

Step 2.5.2

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

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

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 $g (t) = 2 t - 5$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 4

Simplify with factoring out.

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Step 4.1

Multiply $9$ by $- 1$ .

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

Step 4.2

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

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Step 4.2.1

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

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

Step 4.2.2

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

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

Step 4.2.3

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

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

Step 5

Cancel the common factors.

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Step 5.1

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

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

Step 5.2

Cancel the common factor.

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

Step 5.3

Rewrite the expression.

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

Step 6

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

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

Step 7

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

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

Step 8

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

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

Step 9

Multiply $2$ by $1$ .

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

Step 10

Since $- 5$ is constant with respect to $t$ , the derivative of $- 5$ with respect to $t$ is $0$ .

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

Step 11

Simplify the expression.

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Step 11.1

Add $2$ and $0$ .

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

Step 11.2

Multiply $2$ by $- 9$ .

$\frac{2 (2 t - 5) t - 18 (t^{2} + 2)}{{(2 t - 5)}^{10}}$

Step 12

Simplify.

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Step 12.1

Apply the distributive property.

$\frac{(2 (2 t) + 2 \cdot - 5) t - 18 (t^{2} + 2)}{{(2 t - 5)}^{10}}$

Step 12.2

Apply the distributive property.

$\frac{2 (2 t) t + 2 \cdot - 5 t - 18 (t^{2} + 2)}{{(2 t - 5)}^{10}}$

Step 12.3

Apply the distributive property.

$\frac{2 (2 t) t + 2 \cdot - 5 t - 18 t^{2} - 18 \cdot 2}{{(2 t - 5)}^{10}}$

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 $t$ .

$\frac{2 (2 (t \cdot t)) + 2 \cdot - 5 t - 18 t^{2} - 18 \cdot 2}{{(2 t - 5)}^{10}}$

Step 12.4.1.1.2

Multiply $t$ by $t$ .

$\frac{2 (2 t^{2}) + 2 \cdot - 5 t - 18 t^{2} - 18 \cdot 2}{{(2 t - 5)}^{10}}$

Step 12.4.1.2

Multiply $2$ by $2$ .

$\frac{4 t^{2} + 2 \cdot - 5 t - 18 t^{2} - 18 \cdot 2}{{(2 t - 5)}^{10}}$

Step 12.4.1.3

Multiply $2$ by $- 5$ .

$\frac{4 t^{2} - 10 t - 18 t^{2} - 18 \cdot 2}{{(2 t - 5)}^{10}}$

Step 12.4.1.4

Multiply $- 18$ by $2$ .

$\frac{4 t^{2} - 10 t - 18 t^{2} - 36}{{(2 t - 5)}^{10}}$

Step 12.4.2

Subtract $18 t^{2}$ from $4 t^{2}$ .

$\frac{- 14 t^{2} - 10 t - 36}{{(2 t - 5)}^{10}}$

Step 12.5

Factor $2$ out of $- 14 t^{2} - 10 t - 36$ .

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Step 12.5.1

Factor $2$ out of $- 14 t^{2}$ .

$\frac{2 (- 7 t^{2}) - 10 t - 36}{{(2 t - 5)}^{10}}$

Step 12.5.2

Factor $2$ out of $- 10 t$ .

$\frac{2 (- 7 t^{2}) + 2 (- 5 t) - 36}{{(2 t - 5)}^{10}}$

Step 12.5.3

Factor $2$ out of $- 36$ .

$\frac{2 (- 7 t^{2}) + 2 (- 5 t) + 2 (- 18)}{{(2 t - 5)}^{10}}$

Step 12.5.4

Factor $2$ out of $2 (- 7 t^{2}) + 2 (- 5 t)$ .

$\frac{2 (- 7 t^{2} - 5 t) + 2 (- 18)}{{(2 t - 5)}^{10}}$

Step 12.5.5

Factor $2$ out of $2 (- 7 t^{2} - 5 t) + 2 (- 18)$ .

$\frac{2 (- 7 t^{2} - 5 t - 18)}{{(2 t - 5)}^{10}}$

Step 12.6

Factor $- 1$ out of $- 7 t^{2}$ .

$\frac{2 (- (7 t^{2}) - 5 t - 18)}{{(2 t - 5)}^{10}}$

Step 12.7

Factor $- 1$ out of $- 5 t$ .

$\frac{2 (- (7 t^{2}) - (5 t) - 18)}{{(2 t - 5)}^{10}}$

Step 12.8

Factor $- 1$ out of $- (7 t^{2}) - (5 t)$ .

$\frac{2 (- (7 t^{2} + 5 t) - 18)}{{(2 t - 5)}^{10}}$

Step 12.9

Rewrite $- 18$ as $- 1 (18)$ .

$\frac{2 (- (7 t^{2} + 5 t) - 1 (18))}{{(2 t - 5)}^{10}}$

Step 12.10

Factor $- 1$ out of $- (7 t^{2} + 5 t) - 1 (18)$ .

$\frac{2 (- (7 t^{2} + 5 t + 18))}{{(2 t - 5)}^{10}}$

Step 12.11

Rewrite $- (7 t^{2} + 5 t + 18)$ as $- 1 (7 t^{2} + 5 t + 18)$ .

$\frac{2 (- 1 (7 t^{2} + 5 t + 18))}{{(2 t - 5)}^{10}}$

Step 12.12

Move the negative in front of the fraction.

$- \frac{2 (7 t^{2} + 5 t + 18)}{{(2 t - 5)}^{10}}$