What is the 57th derivative of y=cos7x?

Expert Answers
ishpiro eNotes educator| Certified Educator

Let's take a few derivatives of y = cos(7x) in order to observe the pattern they will create:

`y' = -7sin(7x)`

`y'' = -7*7cos(7x) = -7^2cos(7x)`

`y''' = -7^2*(-7)sin(7x) = 7^3sin(7x)`

`y^((4)) = 7^3*7cos(7x) = 7^4cos(7x)`

It is apparent that the power of 7 in front of the trigonometric function will be the same as the order of the derivative. The sign (plus or minus) and whether the function is sine or cosine will have the following order:

-sin

-cos

+sin

+ cos

and this pattern will then repeat all over. Since it will repeat every four derivatives, we can determine how derivative of nth order will look like by dividing n by 4 :

if the remainder is 1 (such as first derivative), the derivative will contain - sin

if the remainder is 2, it will contain -cos

if the remainder is 3, it will contain + sin

if there is no remainder(such as original function, or 4th derivative), it will contain +cos.

So, for the 57th derivative, 57 divides by 4 with the remainder 1. Therefore,

`y^((57)) = -7^57sin(7x)` .

The 57th derivative is `-7^57sin(7x)` .

 

Access hundreds of thousands of answers with a free trial.

Start Free Trial
Ask a Question