Let f(x)= (1+cos^3(7x))^3. what is df/dx when x=pi/4

so far I have

df/dx= 3(1+cos(7x)^3)^3 (3cos(7x)^2 (-sin(7x) (7)

and when i plug in pi/4 i get 20.404 (when accurate to one-thousandths)

is this right?

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- Sep 10th 2009, 07:52 PM #1

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- Sep 10th 2009, 10:01 PM #2
Not sure df/dx is right. Think the first power should be a square not a cube (I may be wrong)

And plugging a number in via a calculator to get a numerical answer is not mathematics. What you need to do is remember that $\displaystyle cos (\pi / 4) = \frac {\sqrt 2} 2$ and same for sine and use that.