Determine the 38th derivative of y=sin(2x)

Explanation:

Alright so this process involves more logic and less of knowledge of how to take 38 derivatives of sin(2x) but we will need to first find the pattern.

So lets start taking derivatives:

y=sin(2x)

Using the chain rule we get: cos(2x) x 2 or 2cos(2x)

This is because the derivative of the outside function sin(x) is cos(x) and the derivative of the inside function or 2x is 2.

So we can say:

y' = 2cos(2x)

If we do this again using the chain rule we get:

2(-sin(2x)) x 2 or -4sin(2x)

So y''= -4sin(2x)

Lets keep going:

y''= -4sin(2x)

y'''= -8cos(2x)

y''''= 16sin(2x)

y''''' = 32cos(2x)

Now we can stop because this is the important place in the pattern. Notice that y' and y''''' look very similar in that they have the same function cos(2x) and this is positive times some constant. This is crucial because then we can say that every 4 derivatives we take from this point the pattern will repeat. This is due to the fact that the derivative of sinx is cosx and the derivative of cosx = -sinx (so eventually things will become repetitive).

Lets list out what we have:

y = sin(2x)

y' = 2cos(2x)

y''= -4sin(2x)

y'''= -8cos(2x)

y''''= 16sin(2x)

y''''' = 32cos(2x)

So we have a pattern we go from positive to two negatives then a positive, and then it will repeat again. So what can we say from this?

Well we can then say that ; where d is some constant. This is because, once again, this pattern is repeating every 4 derivatives. This in turn means that this will hold true for any multiples of 4 so we could say:

Now we can follow the pattern to get the 38th derivative. By this logic our 36th derivative will be a sin(2x), then we go to cos(2x) and then a negative sin(2x) by our general pattern. (see below or above for pattern)

dcos(2x)  to -dsin(2x) to -dcos(2x) to dsin(2x)

36th = dsin(2x) (pattern then repeats)

37th = dcos(2x)

38th = -dsin(2x)

So we can conclude that we have

-dsin(2x) where d is something involving the multiplication of 2 which we have to deduce.

So lets take another look at our derivatives, and specifically the constants in front :

y' = 2cos(2x)

y''= -4sin(2x)

y'''= -8cos(2x)

y''''= 16sin(2x)

y''''' = 32cos(2x)

So we can see that for our first derivative we have a 2, for our second we have 4, 3rd is 8, 4th is 16, and 5th being 32. Now this looks a lot like 2^x, where x is the number of our derivative! Because:

2^1 = 2 (Fits the first derivative)

2^2 = 4 (Fits second)

2^3 = 8 (Fits third)

2^4 = 16

2^5 = 32

So then we can conclude that we will get 2^38 on our 38th derivative!

This means our final answer is

-2^38sin(2x)

actually, no it isn't. when i was in junior high school, it was a blast. considering that it was a pre international baccalaureate school.

explanation: