# This is a GRE prep question: What's the derivative of f(x)=\int_x^0\frac{\cos

This is a GRE prep question:
What's the derivative of $$\displaystyle{f{{\left({x}\right)}}}={\int_{{x}}^{{0}}}{\frac{{{\cos{{x}}}{t}}}{{{t}}}}{\left.{d}{t}\right.}$$

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Alrew1959
First Foundamental Theorem of Calculus :
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\int_{{\alpha{\left({x}\right)}}}^{{\beta{\left({x}\right)}}}}{f{{\left({x},{y}\right)}}}{\left.{d}{y}\right.}={\int_{{\alpha{\left({x}\right)}}}^{{\beta{\left({x}\right)}}}}{\frac{{{d}{f{{\left({x},{y}\right)}}}}}{{{\left.{d}{x}\right.}}}}{\left.{d}{y}\right.}+{\frac{{{d}\beta{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}{f{{\left({x},\beta{\left({x}\right)}\right)}}}-{\frac{{{d}\alpha{\left({x}\right)}}}{{{\left.{d}{x}\right.}}}}{f{{\left({x},\alpha{\left({x}\right)}\right)}}}$$
where:
$$\displaystyle\alpha{\left({x}\right)}={0},\beta{\left({x}\right)}={x},{f{{\left({x},{y}\right)}}}={\frac{{{\cos{{x}}}{y}}}{{{y}}}}$$
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William Yazzie

Note that not only the domain of integration but also the integrand depend on x. Let's (for a moment) write $$F(X)=\int_x^0 f(x,t)dt=-\int_0^x f(x,t)dt$$ for your integral. To handle the "double" x-dependence, we write F as the composition of $$\triangle:\mathbb{R}\to\mathbb{R}^2,x\to (x,x)$$
$$\displaystyle{F}'{\left({x}\right)}=\triangle\Phi{\left(\triangle{\left({x}\right)}\right)}\cdot\triangle'{\left({x}\right)}$$
by the chain rule. Now $$\displaystyle\triangle'{\left({x}\right)}={\left({1},{1}\right)}$$ and
$$\displaystyle{d}_{{1}}\Phi{\left({x}_{{1}},{x}_{{2}}\right)}=-{f{{\left({x}_{{2}},{x}_{{1}}\right)}}}$$
$$\displaystyle{d}_{{2}}\Phi{\left({x}_{{1}},{x}_{{2}}\right)}=-{\int_{{0}}^{{{x}_{{1}}}}}{d}_{{{x}_{{2}}}}{f{{\left({x}_{{2}},{t}\right)}}}{\left.{d}{t}\right.}$$
(for the first derivative we used the fundamental theorem). Plugin everything together, we obtain
$$\displaystyle{F}'{\left({x}\right)}=-{f{{\left({x},{x}\right)}}}-{\int_{{0}}^{{x}}}{d}_{{x}}{f{{\left({x},{t}\right)}}}{\left.{d}{t}\right.}$$