Find $\frac{dy}{dx}$ ($y$ given in terms of a definite integral)

Question

$$y(x) = \displaystyle \int_x^2\frac{1}{t^3} dt$$ Find $\frac{dy}{dx}$

This is the first time I've seen a function defined like that. The first idea that springs to my mind is to evaluate this integral and therefore get this: $$y(x) = - \frac{1}{8}+\frac{1}{2x^2}$$ And now - simply - differentate this function: $$\frac{dy}{dx} = -\frac{1}{x^3} $$ However, I feel really strange about this result - could you tell me if my solution is OK?

Answer 1

if you compute the first derivative of $$\frac{1}{8}-\frac{1}{2x^3}$$ you will get $$-\frac{1}{x^3}$$

Answer 2

Another approach would be to consider that:

$$ y(x) = F(2) - F(x)$$ where F is a primitive function of $ \frac{1}{x^3}$. This means that $ \frac{dy}{dx} = 0 - \frac{dF}{dx} = -\frac{1}{x^3} $ by definition. Keep in mind that $y(x)$ is well defined only if $x > 0$.

Answer 3

The quickest way I can think of is by noticing that

$$-y(x) = \int_2^x \frac{1}{t^3}dt $$

So that by the fundamental theorem of calculus, you get

$$-y'(x) = \frac{1}{x^3} $$