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3 years ago

Find f'(x) for f(x) = cos^2(3x^3).

1 answer:

6 0

$f(x) = \cos^2(3x^3)\\f'(x) = \frac{d}{dx}[\cos^2(3x^3)]\\= 2\cos(3x^3) \cdot \frac{d}{dx}[cos(3x^3)]\\= 2\cos(3x^3) \cdot (-\sin(3x^3)) \cdot \frac{d}{dx}[3x^3]\\= 2\cos(3x^3) \cdot (-\sin(3x^3)) \cdot 9x^2\\= -18x^2\cos(3x^3)\sin(3x^3)$

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Twice the difference of a number and four is less than the sum of the number and five.

Whitepunk [10]

Let the number be x.

2*(x-4) < (x+5).

2x -8 < x +5
2x-x < 5 +8
x<13.

The number is less than 13.

That's it. Cheers.

7 0

3 years ago

67000000 in scientific notation

yan [13]

Answer:

6.7 * 10^7

Step-by-step explanation:

7 0

4 years ago

Pls help im pretty sure its easy i just forgot

dalvyx [7]

the average can be calculated by adding the scores together and then dividing by the number of scores.

we can set up an equation:

let x = test score needed on next test

(72+72+80+x)/4 = 71

multiply both sides by 4

72+72+80+x = 284

add like terms

224+x=284

subtract 224 from both sides

x=60

she will need a 60 for her average to be 71

7 0

3 years ago

Read 2 more answers

(2/3q – 3/4) and (–1/6q – r) What is the coefficient of q in the sum of these two expressions?

oee [108]

Answer:

The co-efficient of q in sum of the given expression is 2

Step-by-step explanation:

Given expressions are $\frac {2}{3q}-\frac {3}{4}$ and $\frac{-1}{6q}-r$

Now sum the given expression

$\frac {2}{3q}-\frac {3}{4} + \frac{-1}{6q} -r = \frac{4-1}{6q} -\frac{3}{4} -r$

$\frac {2}{3q}-\frac {3}{4} + \frac{-1}{6q} -r = \frac{3}{6q} -\frac{3}{4} -r$

$\frac {2}{3q}-\frac {3}{4} + \frac{-1}{6q}-r =\frac{1}{2q} - \frac{3}{4} -r$

Here the co-efficient q is 2 (since $\frac{1}{q} = \frac{1}{2}$ )

3 0

4 years ago

PLEASE HELP ASAP (8)/(27) a^(15) cube of a monomial

vagabundo [1.1K]

Answer:

2/3a^5

Step-by-step explanation:

we know that 2/7x2/7x2/7 = 8/27

a^15 cube will be a^5

because we know (2^3)^2 = 2^6

so (a^5)^3 = a^15

2/3a^5

4 0

3 years ago