# slope of tangent line to a curve

• Mar 11th 2011, 11:00 AM
elprodigioso
slope of tangent line to a curve
ok so here's my question
Find the slope of the tangent line to the curve x^y = y^x at the point (2,4)

im not sure exactly how to go about this, what i've tried so far is taking the natural log of both sides so move the exponents
lnx^y=lny^x
ylnx = xlny
and that's where i get stuck. Am I supposed to differentiate both sides then? Because when I did that so that then I could isolate y' i got the following

y' = y(y-xlny)/x(x-ylnx)
• Mar 11th 2011, 11:04 AM
TheChaz
Good! Now just put 2 in for x, and 4 in for y.
• Mar 11th 2011, 11:10 AM
elprodigioso
Ok so after doing that what do I do? I'm not allowed to use a calculator on this problem so I can't get an exact number, would I leave it simply as
y' = 4(4-2ln4)/2(2-4ln2)
• Mar 11th 2011, 11:17 AM
TheChaz
First, let me disclaim: I haven't actually calculated anything - I'm just helping you with the concept under the assumption that you know the product and chain rules!

The fraction you have can be reduced by 2 on top and bottom. Other than that... I can't come up with any reasonable ways to rewrite. Then again, I'm on my iPhone... ;)
• Mar 11th 2011, 11:32 AM
elprodigioso
Haha that's alright I am SUPPOSED to know how to do this already, but usually we get easier problems like y = x^2-3x or something like that.... i know that for that you would plug in the x in order to get f(a) then take the derivative and set it equal to zero which gives the slope m, then you just use point slope formula y - f(a) = m (x - a)

i'm just having trouble applying that concept to this problem primarily because I don't have it written in terms of y = something you know?
and i don't know if i'm supposed to actually set that derivative that i got, assuming that it's correct, and setting it equal to zero. I know that for this problem a zero would occur where the numerator is zero, so should I set the numerator equal to zero an solve? Mmmmmm I'm just confusing myself :/
• Mar 11th 2011, 06:34 PM
Prove It
Why would you set the derivative equal to 0? You're not evaluating stationary points.

You want to know what the derivative is at the point $\displaystyle (x,y) = (2, 4)$.

You have an $\displaystyle x,y$ relationship to give you the derivative at any point, so substitute $\displaystyle x = 2$ and $\displaystyle yt = 4$.

Once you have the derivative at that point, you can use it and the $\displaystyle (x,y)$ co-ordinate to find the equation of the tangent line using $\displaystyle y = mx + c$...
• Mar 11th 2011, 07:08 PM
elprodigioso
Are you asking me why I would set it equal to zero in general, or why would I set it equal to zero in this particular case? In this case I didn't know if I should as I am not familiar with this problem, and my unfamiliarity with it would be the reason I would attempt it. Either way I was thinking too much about the problem. Question asks for slope, which I have found, it says nothing about the equation of a line tangent at that point.
• Mar 11th 2011, 09:11 PM
NOX Andrew
You found $y' = \dfrac{4(4-2\ln4)}{2(2-4\ln2)}$, which is the slope of the line tangent to the curve at the point (2,4). Here is how I would simplify the answer:

$y' = \dfrac{4(4-2\ln(2^2))}{2(2-4\ln2)}$

$\implies y' = \dfrac{4(4-4\ln2)}{2(2-4\ln2)}$

$\implies y' = \dfrac{16(1-\ln2)}{4(1-2\ln2)}$

$\implies y' = \dfrac{4(1-\ln2)}{(1-2\ln2)}$

If you aren't allowed to use a calculator, then I assume the above answer is acceptable. Any decimal approximation would be just that - an approximation.