Find f'(0), f"(0) : f(x) = xcosx

[Marcia]

This is how i started it.

f'(x) = x(-sinx)
f'(0) = ?????

 

[racuna]

First of all, your differentiation is wrong. Since there is an x in front of cosx, you must apply the product rule. (First times der. of second + second times der. of first) In this case, it would be x(-sinx)+cosx(1). So you simplify and this is f', which is the first derivative. To get f'(0) you just have to plug-in 0 for every x in the funtion.

To get f''(0), you take the derivative of the derivative. (You would also apply the product rule in this case.) When you get your derivative, you can then plug-in the 0 for every x in the equation to evaluate f''(0).
Hope this helps!

-Rachel

 

[soroban]

Hello, Marcia!

Please don't put the problem in the subject line.

Given: f(x) = x·cos(x). . Find: .f '(0), f ''(0)

We must use the Product Rule . . .

f '(x) .= . -x·sin(x) + cos(x)

. . Then: .f '(0) .= .-0·sin(0) + cos(0) .= .1


f ''(x) .= .-x·cos(x) - 1·sin(x) - sin(x) .= .-x·cos(x) - 2·sin(x)

. . Then: .f ''(0) .= .-0·cos(0) - 2·sin(0) .= .0