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3 years ago

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1 answer:

Travka [436]3 years ago

6 0

Answer:

Are we human, or are we dancer

Step-by-step explanation: this is an answer your teacher would love trust me ;)

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If f(x) = 9x10 tan−1x, find f '(x).

djverab [1.8K]

Answer:

$\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

$\displaystyle f(x) = 9x^{10} \tan^{-1}(x)$

<u>Step 2: Differentiate</u>

[Function] Derivative Rule [Product Rule]: $\displaystyle f'(x) = \frac{d}{dx}[9x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Rewrite [Derivative Property - Multiplied Constant]: $\displaystyle f'(x) = 9 \frac{d}{dx}[x^{10}] \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Basic Power Rule: $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + 9x^{10} \frac{d}{dx}[\tan^{-1}(x)]$
Arctrig Derivative: $\displaystyle f'(x) = 90x^9 \tan^{-1}(x) + \frac{9x^{10}}{x^2 + 1}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

7 0

3 years ago

PLEASE HELP ME!!! The sum of two numbers is 46. When 30 is subtracted from 5 times the smaller the result is 3 times the larger.

Ipatiy [6.2K]

Answer:

m = 21

n= 25

Step-by-step explanation:

the equations are:

n + m= 46

5m-30 = 3n

plz vote brainliest :)

8 0

3 years ago

sara drove 130 in 2.5 hours wich equation can be used to couculate r the average speed at which sara drove

goblinko [34]

130 divided by 2.5 or 130/2.5
Answer- 52 mph

3 0

3 years ago

Fifty five miles per hour converted to feet per second

stiks02 [169]

$\frac{55miles}{hour}*\frac{hour}{3600 seconds}*\frac{5280feet}{mile}$
$\frac{55}{1}*\frac{1}{3600 seconds}*\frac{5280feet}{1}=80.67ft/sec$

6 0

3 years ago

Eliminate the parameter. x = t^2 + 1, y = t^2 - 1.

Maksim231197 [3]

Answer:

Eliminating the parameter, the equation is $y = x - 2$

Step-by-step explanation:

We are given the following parametric equations:

$x = t^2 + 1$

$y = t^2 - 1$

We want to eliminate the parameter t. From the first equation:

$x = t^2 + 1$

$t^2 = x - 1$

Replacing in the second equation:

$y = x - 1 - 1$

$y = x - 2$

Eliminating the parameter, the equation is $y = x - 2$

7 0

3 years ago