Trig Derivative

[Denomination]

Find the second derivative of f when

f(x)=cos2x-3(sinx)^2

Answer Choices:

f''(x)= 10cos2x

f''(x)=5cos2x

f''(x)=-10cos2x

f"(x)=-5cos2x

f"(x)=-10sin2x

f"(x)=-5sin2x

I'm writing the answers because I know there is a twist to the way you have to simplify the equation before you solve it but I can't figure out what it is.

 

[galactus]

Did you get:

\(\displaystyle \L\\-4cos(2x)-12cos^{2}(x)+6\) for your 2nd derivative?.

You must use identities to simplify, along with a little factoring.

\(\displaystyle \L\\-4cos(2x)+6\underbrace{(1-2cos^{2}(x))}_{\text{-cos(2x)}}\)


See it now?.

 

[soroban]

Hello, Denomination!

Find the second derivative of f when: \(\displaystyle \,f(x)\:=\:\cos2x\,-\,3\sin^2x\)

Answer Choices:

\(\displaystyle (1)\;10\cos2x\;\;\;(2)\:5\cos2x\;\;\;(3)\;-10\cos2x\;\;\;(4)\;-5\cos2x\;\;\;(5)-10\sin2x\;\;\;(6)\;-5\sin2x\)

I know you have to simplify the equation before you solve it . . . Right!

Use the identity: \(\displaystyle \,\sin^2\theta \:=\:\frac{1\,-\,\cos2\theta}{2}\)

and we have: \(\displaystyle f(x) \:=\:\cos2x\,-\,3\,\sin^2x\:=\:\cos2x\,-\,3\left(\frac{1\,-\,\cos2x}{2}\right)\:=\:\frac{5}{2}\,\cos2x\,-\,\frac{3}{2}\)

Then: \(\displaystyle \:f'(x)\:=\:\frac{5}{2}[-\sin2x](2)\,-\,0\:=\:-5\,\sin2x\)

Therefore: \(\displaystyle \:f''(x)\:=\:-5(\cos2x)(2)\:=\:-10\,\cos2x\) . . . answer choice (c)