How to find all possible functions with the given derivative ? If y′=sin(7t), then y = If y′=cos(t/7), then y = If y′=sin(7t)+cos(t/7), then y =

1 Answer
Sep 3, 2015

#y=-1/7cos(7t)+C#, #y=1/7sin(7t)+C# #y=-1/7cos(7t)+1/7sin(7t)+C#

Explanation:

We know that the derivative (w.r.t. #t#) of #cost# is #-sint#

Using the chain rule, the derivative (w.r.t.#t#) of #cosu# is #-sinu (du)/dt#

So #d/dt(cos(7t)) = -sin(7t) * 7#

If we multiply by the constant #-1/7# before differentiating, we will multiply the derivative by the same constant:

#d/dt(-1/7 cos(7t)) = -1/7(-sin(7t) * 7) = sin(7t)#

So one possible function with derivative #y' = sin(7t)# is

#y = -1/7 cos(7t)#

But there are others.

#y = -1/7 cos(7t) + 7#,
#y = -1/7 cos(7t)-5#,
#y = -1/7 cos(7t)+pi/sqrt17#

Indeed, For any (every) constant #C#, the derivative of #y = -1/7 cos(7t) +C# is the desired derivative.

Not only that, but due to an important consequence of the Mean Value Theorem, every function that has this derivativs differs from #y = -1/7 cos(7t)# by a constant #C#.

Similar reasoning leads us to the functions ahose derivative is #y' = cos(7t)# being expressible as #y=1/7sin(7t)+C# for constant #C#.

Because the derivative of a sum is the sum of the derivatives, every function whose derivative is #y' = sin(7t)+cos(7t)# can be written in the form:

#y = -1/7cos(7t)+1/7sin(7t)+C# for some constant #C#.