derivative..

**BadMonkey** · April 11th 2010, 10:51 AM

if f(x) = (sin^2)x how can I find f''(x) ?

**huraduy** · April 11th 2010, 10:58 AM

Originally Posted by BadMonkey

if f(x) = (sin^2)x how can I find f''(x) ?

f(x)=(sin^2)x=(sinx)^2
f'(x)= 2 (sinx)(cosx)

can you manage the second derivative now?

**Redding1234** · April 11th 2010, 11:01 AM

You will need to use the chain rule for f'(x) and then the product rule for f''(x).

**e^(i*pi)** · April 11th 2010, 11:21 AM

For the second part you can always use the double angle identity for sin: $f'(x) = \sin(2x)$

Then only the chain rule is required

**huraduy** · April 11th 2010, 11:26 AM

Originally Posted by BadMonkey

if f(x) = (sin^2)x how can I find f''(x) ?

now on the second derivative you need to use the product rule, after that apply the trig identity. My result for that problem is -2.

**e^(i*pi)** · April 11th 2010, 11:52 AM

Originally Posted by huraduy

now on the second derivative you need to use the product rule, after that apply the trig identity. My result for that problem is -2.

I get $f''(x) = -2\cos(2x)$

**CrashDummy11** · April 11th 2010, 12:31 PM

I always use Calc101.com Automatic Calculus, Linear Algebra, and Polynomials when there are derivatives that I am not sure of. Should be able to help with yours.

**huraduy** · April 11th 2010, 04:31 PM

Originally Posted by e^(i*pi)

I get $f''(x) = -2\cos(2x)$

I am sorry, you are right. I forgot my negative sign.

take the -2((sin^2)x-(cos^2)x)=-2cos2x.

Thank you for letting me know

Thread: derivative..