I've been asked to find the derivative of $$g(x)= \int_{\cos x}^{x^4}\sqrt{2-u} du$$ using the Fundamental Theorem of Calculus part 1, and I know I should be substituting and setting a variable to one of the bounds, but I'm not sure how to tackle this with both bounds being functions. Any advice is incredibly appreciated.
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$\begingroup$The fundamental theorem of calculus says that $$g(x)=\frac d {dx}\,\int_{a(x)}^{b(x)}f(u)\,du=f(b(x))\, b'(x)-f(a(x))\, a'(x)$$ In your case $$f(u)=\sqrt{2-u}\quad,\quad a(x)=\cos(x) \quad,\quad b(x)=x^4$$ So, just apply.
If the presence of two bounds makes a problem to you, just consider that $$\int_{a(x)}^{b(x)} =\int_{a(x)}^{0}+\int_{0}^{b(x)}=\int_{0}^{b(x)}-\int_0^{a(x)}$$
$\endgroup$ 5 $\begingroup$By the fundamental theorem,
$$g(x)= \int_{\cos(x)}^{x^4}\sqrt{2-u}\,du=\left.F(u)\right|_{\cos(x)}^{x^4}=F(x^4)-F(\cos(x))$$ where $F(u)$ is the antiderivative of $f(u)=\sqrt{2-u} $.
Then, by the chain rule,
$$g'(x)=4x^3f(x^4)-\sin(x)f(\cos(x))=4x^3\sqrt{2-x^4}-\sin(x)\sqrt{2-\cos(x)}$$
$\endgroup$ $\begingroup$Applying Fundamental Theorem of Calculus, one should get $$g(x)=\int_{\cos x}^{x^4}\sqrt{2-u}\ du$$ $$g'(x)=\sqrt{2-x^4}\cdot \frac{d}{dx}(x^4)-\sqrt{2-\cos x}\cdot \frac{d}{dx}(\cos x)$$ $$=\sqrt{2-x^4}\cdot (4x^3)-\sqrt{2-\cos x}\cdot (-\sin x)$$ $$=4x^3\sqrt{2-x^4}+\sin x\sqrt{2-\cos x}$$
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