# calculate the derivative using fundamental theorem of calculus

This is a GRE prep question: What's the derivative of $f(x)=\int_x^0 \frac{\cos xt}{t}\mathrm{d}t$?

The answer is $\frac{1}{x}[1-2\cos x^2]$. I guess this has something to do with the first fundamental theorem of calculus but I'm not sure how to use that to solve this problem.

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## migrated from mathoverflow.netJul 5 '13 at 6:00

This question came from our site for professional mathematicians.

$\int_a^b=-\int_b^a$ –  Babak S. Jul 5 '13 at 6:03

First Foundamental Theorem of Calculus : $$\frac{d}{dx}\int_{{\alpha(x)}}^{{\beta(x)}}f(x,y)\,dy=\int_{{\alpha(x)}}^{{% \beta(x)}}\frac{\partial f(x,y)}{\partial x}\,dy+\frac{d\beta(x)}{dx}f(x,\beta% (x))-\frac{d\alpha(x)}{dx}f(x,\alpha(x))\,$$ where : $$\alpha(x)=0, \beta(x)=x, f(x,y)=\frac{\cos xy}{y}$$

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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 $\Delta\colon \mathbb R \to \mathbb R^2$, $x\mapsto (x,x)$ and $\Phi\colon \mathbb R^2 \to \mathbb R$, $(x_1, x_2) \mapsto -\int_0^{x_1} f(x_2, t)\, dt$. We have $F = \Phi \circ \Delta$, hence $$F'(x) = \nabla \Phi \bigl(\Delta(x)\bigr) \cdot \Delta'(x)$$ by the chain rule. Now $\Delta'(x) = (1,1)$, and \begin{align*} \partial_1\Phi(x_1, x_2) &= -f(x_2, x_1)\\ \partial_2\Phi(x_1, x_2) &= -\int_0^{x_1} \partial_{x_2} f(x_2, t)\, dt \end{align*} (for the first derivative we used the fundamental theorem). Plugin everything together, we obtain $$F'(x) = -f(x,x) - \int_0^x \partial_x f(x,t)\, dt$$

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The integral does not exist, consequently it is not differentiable.

The integral does not exist, because for each $x$ ($x>0$, the case of negative $x$ is dealt with similarly) there is some $\varepsilon > 0$ such that $\cos(xt)> 1/2$ for all $t$ in the $t$-range $0\le t \le \varepsilon$. If one splits the integral $\int_x^0 = \int_\varepsilon^0 + \int_x^\varepsilon$, then the second integral exists, because the integrand is continuous in the $t$-range $\varepsilon \le t\le x$. The first is by definition $\lim_{\eta\rightarrow 0+}\int_\varepsilon^\eta$ and larger than $(-\ln \eta + \ln \varepsilon)/2$. So the limit does not exist.

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