# Thread: Differentiation problem (from GRE math subject test)

1. ## Differentiation problem (from GRE math subject test)

Hi everyone,

I, for the life of me, can't seem to figure this one out. It's from the GRE math subject test.

Let $f(x) = \int_{0}^{x} \frac{cos(xt)}{t}dt$.

Then, find $f'(x)$

If we could pull out the $x$ from the integral, then it's just product rule and FTC. But that $x$ is tied up in there.

Any ideas from the calculus geniuses out there?

2. Originally Posted by GnomeSain
Hi everyone,

I, for the life of me, can't seem to figure this one out. It's from the GRE math subject test.

Let $f(x) = \int_{0}^{x} \frac{cos(xt)}{t}dt$.

Then, find $f'(x)$

If we could pull out the $x$ from the integral, then it's just product rule and FTC. But that $x$ is tied up in there.

Any ideas from the calculus geniuses out there?
Fundamental Theorem of Calculus ...

if $\displaystyle f(x) = \int_a^x g(t) \, dt$ , then $f'(x) = g(x)$

3. So notice that your integrand is just a function of t, $g(t)$. But does the FTC still apply when we have $g(x,t)$ which is the case in my initial problem?

Originally Posted by skeeter
Fundamental Theorem of Calculus ...

if $\displaystyle f(x) = \int_a^x g(t) \, dt$ , then $f'(x) = g(x)$

4. Originally Posted by GnomeSain
So notice that your integrand is just a function of t, $g(t)$. But does the FTC still apply when we have $g(x,t)$ which is the case in my initial problem?
t is a dummy variable ... still applies.

5. Of course [LaTeX ERROR: Convert failed] applies. That's just the fundamental theorem of calculus. What I've been asking all along is what about [LaTeX ERROR: Convert failed] ?

Again, the issue here is the integrand is a function of both [LaTeX ERROR: Convert failed] and [LaTeX ERROR: Convert failed] . Furthermore, these variables aren't easily separable.

Originally Posted by skeeter
t is a dummy variable ... still applies.

6. sorry ... did not see the "x" in the integrand.

7. The big formula here may be relevant.

8. Lagrange's formula extends the fundamental theorem of calculus:
$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t) dt= f(x, \beta(x))\frac{d\beta(x)}{dx}- f(x,\alpha(x))\frac{d\alpha(x)}{dx}+ \int_alpha(x)^\beta(x)\frac{\partialf(x,t)}{\parti al x} dt$

Here, $\alpha(x)= 0$ and $\beta(x)= x$ so that becomes
$\frac{cos(xt)}{x}- \int_0^x \frac{sin(xt)}{t}dt$

However, it looks to me like the very existence of this function is questionable. What happens to $\frac{cos(xt)}{t}$ at t= 0, the lower limit of the integral?

9. The "x" inside the cosine makes this a little more than just the fundamental theorem:

Leibniz's formula extends the fundamental theorem of calculus:
$\frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t) dt= f(x, \beta(x))\frac{d\beta(x)}{dx}- f(x,\alpha(x))\frac{d\alpha(x)}{dx}+ \int_alpha(x)^\beta(x)\frac{\partialf(x,t)}{\parti al x} dt$

Here, $\alpha(x)= 0$ and $\beta(x)= x$ so that becomes
$\frac{cos(xt)}{x}- \int_0^x \frac{sin(xt)}{t}dt$
(Ah, I see that is the same formula that Ackbeet linked to.)

However, it looks to me like the very existence of this function is questionable. What happens to $\frac{cos(xt)}{t}$ at t= 0, the lower limit of the integral?