derivatives and such

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whiteti

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Jun 3, 2013
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find the equation of the tangent line to the graph of f(x)=8sin^3xcosx at x=7pi/4
f'(x)=8(3)sin^2 xcosx
=24cos^2 x * cosx - 8sin^3x * sinx
=24cos^2 x - 8sin^4 x

(the x's are not part of the powers)

now im stuck... I know its taylor expansion but I dont know how to put them together. please help!
 
whiteti said:
find the equation of the tangent line to the graph of f(x)=8sin^3xcosx at x=7pi/4
f'(x)=8(3)sin^2 xcosx + . . .
=24cos^2 x * cosx - 8sin^3x * sinx
=24cos^2 x - 8sin^4 x

(the x's are not part of the powers)

now im stuck... I know its taylor expansion but I dont know how to put them together. please help!
Click to expand...
f(x) = 8 sin^3(x) cos(x)

This is a product of two factors, so you have to use the product rule:
Your derivative has to have two terms, one with 8sin^3(x) differentiated, and the other with cos(x) differentiated. Further, the derivative of the first term needs the chain rule .. that may be what you have left out. You DO need to check your work! You also might find trig identities that will simplify the form.

After you have the derivative, then substitute x = 7pi/4 OR equivalently, x = -pi/4.
 
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