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September 22, 2014

September 22, 2014

Posted by **stacy** on Monday, February 27, 2012 at 12:21am.

2. If d/dx(f(3x^3))=9x^2, Calculate f'(x).

3. Let f(x)=cos(6x+6), find f'(x)

4. Let f(x)=-2sec(2x), find f'(x)

- COLLEGE CALCULUS. HELP! -
**Steve**, Monday, February 27, 2012 at 12:03pmLooks like an exercise in the chain rule:

f(x) = sin(x^2)

f'(x) = cos(x^2) * 2x = 2x cos(x^2)

f'(1) = 2cos(1)

--------------------

d/dx(f(3x^3))=9x^2

Let g(x) = f(3x^3)

dg/dx = dg/df * df/dx

9x^2 = dg/df * 9x^2

1 = dg/df

Looks like g(x) = x

so, g(f) = f

dg/dx = df/dx = 9x^2

--------------------

f(x)=cos(6x+6)

f' = -sin(6x+6)*6

= -6sin(6x+6)

----------------

f(x)=-2sec(2x)

f' = -2sec(2x)tan(2x)*2

= -2sec(2x)tan(2x)

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