Is my solution of $\frac{d }{dx}\int_0^{\cos x}\sqrt{1+t^4}dt$ correct? $$\frac{d }{dx}\int_0^{\cos x}\sqrt{1+t^4}dt$$
$$\frac{(\sqrt{1+\cos^4x}-1)dx}{dx}$$
$$\sqrt{1+\cos^4x}-1$$
The answer seems weird to me, but I see no other way to do this. Was this correct? If not, how to do it?
Thank you.
 A: No, sorry. Differentiating with respect to $\cos x$ gives $\sqrt{1+\cos^4 x}$. Differentiating with respect to $x$ instead gives $-\sqrt{1+\cos^4 x}\sin x$.
A: $\int_0^{\cos x}\sqrt{1+t^4}dt=F\circ G$ where $F(x)=\int_0^{ x}\sqrt{1+t^4}dt $ and $G(x)=cos(x)$ apply $(F\circ G)'(x)=F'(G(x)).G'(x)$
$-sin(x)\sqrt{1+cos^4(x)}$.
A: Let an antiderivative of the integrand $f(t)$ be $F(t)$. Then by the fundamental theorem of calculus
$$\frac{d }{dx}\int_0^{\cos x}\sqrt{1+t^4}\,dt=\frac d{dx}(F(\cos x)-F(0))=f(\cos x)(-\sin x)=(-\sin x)\sqrt{1+\cos^4t}$$
A: You forgot to multiply by the derivative of $\cos(x)$ and the $1$ at the end will not come. The correct answer is: 
$$\sqrt{1+\cos^4(x)} \times -\sin(x)$$
A: All of the answers are correct, but do not really explain why.
$$\dfrac{d}{dx}\left(\int_{f(x)}^{g(x)} h(t)dt\right)$$
Let $H(t)$ be any antiderivative of $h(t)$. In other words, $\dfrac{d}{dt}H(t) = h(t)$.
Now, we can apply the Fundamental Theorem of Calculus to the integral:
$$\dfrac{d}{dx}\left(\int_{f(x)}^{g(x)}h(t)dt\right) = \dfrac{d}{dx}\left(H(g(x))-H(f(x)) \right)$$
Next, we apply the Chain Rule:
$$\dfrac{d}{dx}\left(H(g(x)) - H(f(x))\right) = H'(g(x))g'(x)-H'(f(x))f'(x) = h(g(x))g'(x)-h(f(x))f'(x)$$
For your integral, you have $h(t) = \sqrt{1+t^4}$, $g(x) = \cos x$, and $f(x)=0$. So, as others have pointed out, this becomes:
$$\sqrt{1+\cos^4 x}\left(-\sin x\right) - 0 = -\sin x \sqrt{1+\cos^4 x}$$
Edit: Actually, Tsemo Aristide's and Parcly Taxel's answers do give a reason, and this answer is an alternate interpretation from Tsemo's and an expansion on Parcly's answer leading to the same result.
