Question

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 =

Answer 1

`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`.