# Thread: Another Derivative of a Sin Function Problem

1. ## Another Derivative of a Sin Function Problem

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!

2. Almost but you're forgetting to differentiate the 'sin' part of sin(x + n) so the answer should be...
$9\cos(x+n) - 2\sin(2x)$

Is it an n or a pi? Won't matter anyway answer would be the same but exchange the n for a pi!

3. sin(x+ pi) ' = cos(x+ pi)

d/dx(9sin(x+ pi) +cos(2x)) = 9cos(x+ pi) - 2sin(2x)

4. 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.

5. Originally Posted by r2d2
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.
that is correct.

6. And finally, the 3rd Derivative!

f'''(x)= -9sin(x+pi) - 4cos2x
f''''(x)= -9(cos(x+pi)) - [4sin(2x)*-2]
f'''(x)= -9cos(x+pi) - [-8sin(2x)]

f'''(x)= -9 cos(x+pi) + 8sin2x

This would be the 3rd derivative, correct?

Thanks will be going around for all!

7. Originally Posted by r2d2
And finally, the 3rd Derivative!

f'''(x)= -9sin(x+pi) - 4cos2x
f''''(x)= -9(cos(x+pi)) - [4sin(2x)*-2]
f'''(x)= -9cos(x+pi) - [-8sin(2x)]

f'''(x)= -9 cos(x+pi) + 8sin2x

This would be the 3rd derivative, correct?

Thanks will be going around for all!
will you require the same reassurance when you take the 99th derivative?

have some confidence in your calculations.

8. Well my BC teacher didn't assign the 99th derivative but if he did, I probably would be checking myself until then!

I definitely understand this derivative of trigonometric functions now. Thanks a lot.