Derivatives of Integrals Analysis Find the derivative of the functions:
$$\int_{x^2}^{\sin(x)}\sqrt{1+t^4}dt$$
In class we had the following solution:
By the fundamental theorem of calculus we know that 
$$\left(\int_a^xf(t)dt\right)'=f(x)$$ So
$$\int_{x^2}^0\sqrt{1+t^4}dt+\int_0^{\sin(x)}\sqrt{1+t^4}dt=$$
$$\int_0^{\sin(x)}\sqrt{1+t^4}dt-\int_0^{x^2}\sqrt{1+t^4}dt=$$
Letting $g(x)=\sqrt{1+t^4} $
$$g(\sin(x))(\sin(x))'-g(x^2)(x^2)'=$$
$$\sqrt{1+\sin(x)^4}\cdot \cos(x)-\sqrt{1+x^8} \cdot 2x$$
However, if we have that $\left(\int_a^xf(t)dt \right)'=f(x)$ wouldn't the answer just be 
$$\sqrt{1+\sin(x)^4}-\sqrt{1+x^8}?$$
 A: You are forgetting the chain rule; you have to take the derivative of each function in the limits.  Thus your derivative is
$$\sqrt{1+\sin^4{x}} \frac{d}{dx}\sin{x} - \sqrt{1+x^8} \frac{d}{dx} x^2$$
which I believes gives you your answer.
A: Generally, the way @Ron walked, is known as Leibniz integral rule which follows from the chain rule:
 $$ {d\over dx} \left( \int_{f(x)}^{g(x)} \phi(t) \,dt \right )= \phi\left(g(x)\right) {g'(x)} -  \phi(f(x)) {f'(x)} $$
A: The function $\,\sqrt{1+t^4}\;$ is defined and continuous everywhere, from where we can thus apply the Fundamental Theorem of Integral Calculus and get
$$\int\limits_{x^2}^{\sin x}\sqrt{1+t^4}dt=F(\sin x)-F(x^2)$$
with $\;F\;$ a function s.t. $\,F'(x)=\sqrt{1+x^4}\;$ (the primitive function of the integrand).
Thus we get, applying the chain rule:
$$\left(\int\limits_{x^2}^{\sin x}\sqrt{1+t^4}dt\right)'=\left(F(\sin x)-F(x^2)\right)'=\cos xF'(\sin x)-2xF'(x^2) =$$
$$=\cos x\sqrt{1+\sin^4x}-2x\sqrt{1+x^8}$$
A: 
However, if we have that $\left(\int_a^xf(t)dt \right)'=f(x)$ wouldn't the answer just be $$\sqrt{1+\sin(x)^4}-\sqrt{1+x^8}?$$

Note that hidden in the first formula you mention is multiplication by $\frac {d}{dx} (x)$. The reason you don't see it is because it is equal to one! It might be more clear to write 
$$\left(\int_a^x f(t)\,dt\right)' = f(x) \frac{d}{dx}(x)$$ 
since the more general formula is 
$$\left(\int_{u(x)}^{v(x)} f(t)\,dt\right)' = f(v(x)) \frac{dv}{dx} - f(u(x))\frac{du}{dx}$$
or even more generally, for $f(x,t)$ continuous,
$$\left(\int_{u(x)}^{v(x)} f(x,t)\,dt\right)' = \int_{u(x)}^{v(x)}\frac{\partial } {\partial x} f(x,t) +  f(x,v(x)) \frac{dv}{dx} - f(x,u(x))\frac{du}{dx}.$$
This last formula easily reduces to the simpler case, since $\frac{\partial } {\partial x} f(t) =0$. 
