if f(x) = (sin^2)x how can I find f''(x) ?
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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?
You will need to use the chain rule for f'(x) and then the product rule for f''(x).
For the second part you can always use the double angle identity for sin: $\displaystyle f'(x) = \sin(2x)$ Then only the chain rule is required
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.
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 $\displaystyle f''(x) = -2\cos(2x)$
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.
Originally Posted by e^(i*pi) I get $\displaystyle f''(x) = -2\cos(2x)$ 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
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