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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Suppose it takes 45 hours for robot A to construct a new robot. Working together with robot B, it takes 25 hours for both robots

Len [333]

Answer:

56 hours 25 minutes

Step-by-step explanation:

Given:

Suppose it takes 45 hours for robot A to construct a new robot

It takes 25 hours for both robots to construct a new robot.

Question asked:

How long would it take robot B to construct a new robot, working alone ?

Solution:

Let the time taken by robot B to construct new robot = $x$

By robot A

It takes 45 hours to construct = 1 new robot

It takes 1 hour to construct = $\frac{1}{45} \ new\ robot$

By robot B

It takes $x$ hours to construct = 1 new robot

It takes 1 hour to construct = $\frac{1}{x}$ new robot

By working together

It takes 25 hours to construct = 1 new robot

It takes 1 hour to construct = $\frac{1}{25} \ new\ robot$

$\frac{1}{25}$ new robot is constructed in = 1 hour

New robot is constructed by both working together in 1 hour = New robot is constructed by robot A in 1 hour + New robot is constructed by robot B in 1 hour

$\frac{1}{25} =\frac{1}{45} +\frac{1}{x} \\$

Subtracting both sides by $\frac{1}{45}$

$\frac{1}{25}-\frac{1}{45} =\frac{1}{45} -\frac{1}{45}+\frac{1}{x} \\\\\frac{1}{25}-\frac{1}{45} =\frac{1}{x}\\\\ Taking\ LCM \ of \ 25\ and\ 45,\ we\ got\ 225$

$\frac{9-5}{225} =\frac{1}{x} \\ \\ \frac{4}{225} =\frac{1}{x}\\\\ By\ cross \ multiplication\\4\times x=225\\Dividing\ both\ sides\ by\ 4\\x=56.25\ hours$