# Differentiation problem (from GRE math subject test)

• Sep 9th 2010, 03:09 PM
GnomeSain
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 $\displaystyle f(x) = \int_{0}^{x} \frac{cos(xt)}{t}dt$.

Then, find $\displaystyle f'(x)$

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

Any ideas from the calculus geniuses out there?
• Sep 9th 2010, 04:27 PM
skeeter
Quote:

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

Then, find $\displaystyle f'(x)$

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

Any ideas from the calculus geniuses out there?

Fundamental Theorem of Calculus ...

if $\displaystyle \displaystyle f(x) = \int_a^x g(t) \, dt$ , then $\displaystyle f'(x) = g(x)$
• Sep 9th 2010, 05:36 PM
GnomeSain
So notice that your integrand is just a function of t, $\displaystyle g(t)$. But does the FTC still apply when we have $\displaystyle g(x,t)$ which is the case in my initial problem?

Quote:

Originally Posted by skeeter
Fundamental Theorem of Calculus ...

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

• Sep 9th 2010, 05:39 PM
skeeter
Quote:

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

t is a dummy variable ... still applies.
• Sep 9th 2010, 10:02 PM
GnomeSain
Of course $\displaystyle t$ applies. That's just the fundamental theorem of calculus. What I've been asking all along is what about $\displaystyle x$?

Again, the issue here is the integrand is a function of both $\displaystyle t$ and $\displaystyle x$. Furthermore, these variables aren't easily separable.

Quote:

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

• Sep 10th 2010, 04:48 AM
skeeter
sorry ... did not see the "x" in the integrand.
• Sep 10th 2010, 04:58 AM
Ackbeet
The big formula here may be relevant.
• Sep 11th 2010, 05:50 AM
HallsofIvy
Lagrange's formula extends the fundamental theorem of calculus:
$\displaystyle \frac{d}{dx}\int_{\alpha(x)}^{\beta(x)} f(x,t) dt= f(x, \beta(x))\frac{d\beta(x)}{dx}- f(x,\alpha(x))\frac{d\alpha(x)}{dx}+ \int_alpha(x)^\beta(x)\frac{\partialf(x,t)}{\parti al x} dt$

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

However, it looks to me like the very existence of this function is questionable. What happens to $\displaystyle \frac{cos(xt)}{t}$ at t= 0, the lower limit of the integral?
• Sep 11th 2010, 05:53 AM
HallsofIvy
The "x" inside the cosine makes this a little more than just the fundamental theorem:

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

Here, $\displaystyle \alpha(x)= 0$ and $\displaystyle \beta(x)= x$ so that becomes
$\displaystyle \frac{cos(xt)}{x}- \int_0^x \frac{sin(xt)}{t}dt$
(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 $\displaystyle \frac{cos(xt)}{t}$ at t= 0, the lower limit of the integral?