Differentiation problem (from GRE math subject test)

**GnomeSain** · September 9th 2010, 03:09 PM

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?

**skeeter** · September 9th 2010, 04:27 PM

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 $f'(x) = g(x)$

**GnomeSain** · September 9th 2010, 05:36 PM

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 $f'(x) = g(x)$

**skeeter** · September 9th 2010, 05:39 PM

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.

**GnomeSain** · September 9th 2010, 10:02 PM

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.

**skeeter** · September 10th 2010, 04:48 AM

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

**Ackbeet** · September 10th 2010, 04:58 AM

The big formula here may be relevant.

**HallsofIvy** · September 11th 2010, 05:50 AM

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?

**HallsofIvy** · September 11th 2010, 05:53 AM

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?

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