Let f(x)= 9sin(x+π) + cos2x
a) Find f'(x)
b) Find f''(x)
c) Find f'''(x)
Would the first derivative be this equation?
f'(x)= 9 * Sin(x+π)*1 + 0*Sin(x+π) + -2sin(2x)?
Would that be correct, using the chain rule, etc?
Thanks!
Let f(x)= 9sin(x+π) + cos2x
a) Find f'(x)
b) Find f''(x)
c) Find f'''(x)
Would the first derivative be this equation?
f'(x)= 9 * Sin(x+π)*1 + 0*Sin(x+π) + -2sin(2x)?
Would that be correct, using the chain rule, etc?
Thanks!
Ok so from there, let's see if the 2nd derivative is correct that I have.
f''(x)= 9cos(x+pi) - 2sin(2x)
f''(x)= 9(-sin(x+pi)) - [2cos(2x)*2]
f''(x)= -9sin(x+pi) - 4cos(2x)
So would the second derivative be:
f''(x)= -9sin(x+pi) - 4cos(2x)
Thanks if anyone could verify that.