Question

Find dy/dx

Answer 1

\[y=\int\limits_{0}^{\sqrt{t}} (\sin t^2) dt\]

Answer 2

Is the upper limit suppose to be \(\large \sqrt{x}\) by chance?

Answer 3

Oh yes sorry about that

Answer 4

We'll use the `Fundamental Theorem of Calculus, Part 1`:\[\large \frac{d}{dx}\int\limits_0^x f(t)dt \qquad = \qquad f(x)\]

The idea is, you're taking the anti-derivative, then you're taking the derivative.
So you're basically undoing whatever you did.

What happens between those two steps is important though.
The function changes due to the limits of integration.

Answer 5

In this problem, we have to be careful with applying the FTC.
Since our upper limit is more than just \(\large x\),
we need to make sure we apply the chain rule.

Answer 6

Grr my head's not working tonight XD wanna trying to come up with an example to try to explain this. The chain rule part can be a lil tricky.

Answer 7

\[\large \frac{d}{dx}\int\limits_1^{x^2} 2t\;dt\]Taking the integral gives us,\[\large \frac{d}{dx}\left[t^2\right]_0^{x^2} \qquad = \qquad \frac{d}{dx}\left[(x^2)^2-1^2\right]\]Taking the derivative of each term gives us,\[\large 2(x^2)\frac{d}{dx}x^2\]

See how our functioned changed from 2t to 2x^2 ?
That's due to the upper limit of integration, that's what becomes the new variable in our expression.

We then apply the chain rule..

Answer 8

Ok ok maybe the example was unnecessary.. Let's just get back to this problem :3

Answer 9

\[\large y\qquad =\qquad \int\limits\limits_{0}^{\color{orangered}{\sqrt{x}}} \sin\left[ \color{orangered}{t}^2\right] dt\]

\[\large \frac{dy}{dx} \qquad = \qquad \sin\left[(\color{orangered}{\sqrt x})^2\right]\frac{d}{dx}\sqrt x\]

Answer 10

so that's the answer?