Derivative of $\int_c^{\sqrt{x}}1\,dt$ - Application of Fundamental Theorem of Calculus

Question

I am trying to determine the $\frac{d}{dx}\int_c^{\sqrt{x}}1\,dt$.

I know that the fundamental theorem states the following, $$\frac{d}{dx}\int_c^x f(t)\,dt=f(t)$$

However in this case the function is a constant, usually if our upper limit of integration is a function itself we end up taking the derivative of the function when applying chain rule as follows, $\frac{d}{dx}\int_c^{g(x)} f(t) \, dt = f(g(x))\cdot g'(x)$

However in this case since the function is a constant can we even apply chain rule? I think the answer should be as follows,

$$\frac{d}{dx}\int_c^{\sqrt{x}}1\,dt= 1$$

But I'm pretty sure the right answer is $\frac{d}{dx}\int_c^{\sqrt{x}}1\,dt= \frac{1}{2\sqrt{x}}$

Could someone explain to me why my answer is wrong?

Answer 1

Let $f(x) = \displaystyle \int_c^{\sqrt{x}} 1\,dt$, $u(x) = \sqrt{x}, g(w) = \displaystyle \int_c^w 1 \,dt$. You have: $f(x)= (g\circ u)(x)\displaystyle \implies g'(u) =1$ by the Fundamental Theorem Of Calculus. By the Chain Rule, you have $f'(x) = g'(u)\cdot u'(x)= 1\cdot \dfrac{1}{2\sqrt{x}}= \dfrac{1}{2\sqrt{x}}$ .