# 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 =

Sep 3, 2015

$y = - \frac{1}{7} \cos \left(7 t\right) + C$, $y = \frac{1}{7} \sin \left(7 t\right) + C$ $y = - \frac{1}{7} \cos \left(7 t\right) + \frac{1}{7} \sin \left(7 t\right) + C$

#### Explanation:

We know that the derivative (w.r.t. $t$) of $\cos t$ is $- \sin t$

Using the chain rule, the derivative (w.r.t.$t$) of $\cos u$ is $- \sin u \frac{\mathrm{du}}{\mathrm{dt}}$

So $\frac{d}{\mathrm{dt}} \left(\cos \left(7 t\right)\right) = - \sin \left(7 t\right) \cdot 7$

If we multiply by the constant $- \frac{1}{7}$ before differentiating, we will multiply the derivative by the same constant:

$\frac{d}{\mathrm{dt}} \left(- \frac{1}{7} \cos \left(7 t\right)\right) = - \frac{1}{7} \left(- \sin \left(7 t\right) \cdot 7\right) = \sin \left(7 t\right)$

So one possible function with derivative $y ' = \sin \left(7 t\right)$ is

$y = - \frac{1}{7} \cos \left(7 t\right)$

But there are others.

$y = - \frac{1}{7} \cos \left(7 t\right) + 7$,
$y = - \frac{1}{7} \cos \left(7 t\right) - 5$,
$y = - \frac{1}{7} \cos \left(7 t\right) + \frac{\pi}{\sqrt{17}}$

Indeed, For any (every) constant $C$, the derivative of $y = - \frac{1}{7} \cos \left(7 t\right) + C$ 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 $y = - \frac{1}{7} \cos \left(7 t\right)$ by a constant $C$.

Similar reasoning leads us to the functions ahose derivative is $y ' = \cos \left(7 t\right)$ being expressible as $y = \frac{1}{7} \sin \left(7 t\right) + C$ for constant $C$.

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

$y = - \frac{1}{7} \cos \left(7 t\right) + \frac{1}{7} \sin \left(7 t\right) + C$ for some constant $C$.