# Thread: Finding the 79th derivative of something

1. ## Finding the 79th derivative of something

Hello everyone, I am currently trying to solve a problem that invloves finding the 79th derivative of something and am trying to look for a pattern, but am not sure if I'm doing this right.

Sin8u,

I've gotten the following:

cosu*8
-64cos8x
-512sin8x
4096sin8x

Any advice on this would be greatly appreciated. I want to know how to do this!

2. If $f(x) = \sin(ax+b)$, then $f^{(n)}(x) = a^n\sin(ax+b+\frac{n\pi}{2})$. You can prove this by induction.

3. I would probably do it this way. Note that

$y(u)=\sin(8u)$

$y'(u)=8\cos(8u)$

$y''(u)=-64\sin(8u)$

$y'''(u)=-512\cos(8u)$

$y^{(4)}(u)=4096\cos(8u).$

Now you can see the pattern: sin, cos, -sin, -cos, and then you repeat. The coefficient just goes as the power of 8. So the question really boils down to this: what is the remainder when you divide 79 by 4?

4. Okay I knew there had to be somekind of patter in this. So, I divided 79 by 4 and got a remainder of 3. So does that mean I pick the third choice in the sequence? So...

Something like 2.20X10^(71)-cose(8U)?

5. Okay I did end up getting it right and understand the whole process, thank you everyone that helped and took the effort. Thank you again!

6. You should write it a bit better. Here is WolframAlpha's answer.

7. Originally Posted by tastybrownies
Hello everyone, I am currently trying to solve a problem that invloves finding the 79th derivative of something and am trying to look for a pattern, but am not sure if I'm doing this right.

Sin8u,

I've gotten the following:

cosu*8
-64cos8x
-512sin8x
4096sin8x

Any advice on this would be greatly appreciated. I want to know how to do this!
$y = sin(8x) = Im\left[e^{8ix}\right]$.

Therefore:

$\frac{d^{79}y}{dx^{79}} = \frac{d^{79}}{dx^{79}} Im\left[e^{8ix}\right] = Im\left[ \frac{d^{79} e^{8ix}}{dx^{79}}\right] = Im\left[ (8i)^{79} e^{8ix} \right]$

$= -8^{79}Im\left[i e^{8ix} \right] = -8^{79} \cos(8x)$.