Derivative of (tan x)^(ln x)?

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csmajor86

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I am working on a homework problem which asks for the derivative of y = (tan x)^ ln x .

My strategy is to take the natural log of both sides which gives me: ln y = ln(x) *ln(tan x) , after bringing down the ln(x). From here I am using implicit differentiation and the "product rule" and then plugging the original (tan x)^ ln x back in for y. This gives me a very large mess which I can't seem to simplify very well and I think I'm doing something wrong. Can anyone help me see where and if I'm going wrong?

Thank you
 
What did you get as a final solution.

Usually with problems dealing with implicit differentiation, you'll end up with very messy results for y' and it would be perfectly acceptable for your solution to contain both variables x and y.
 
I got dy/dx = (y * x ln(x) sec(x) + ln(tan x) sin(x)) / x sin(x) then I substituted tan(x)^ln(x) back in for y so that I get the derivative with respect to x.
 
\(\displaystyle y = (tanx)^{lnx}\)

\(\displaystyle lny = lnx \cdot ln(tanx)\)

\(\displaystyle \frac{y'}{y} = (lnx)' \cdot ln(tanx) + lnx \cdot \left[ln(tanx)\right]' \quad \mbox{Product Rule}\)

I'm not sure how you got secx and sinx involved. Are you sure you got the derivative of tanx right when using the chain rule?:

\(\displaystyle \frac{d}{dx} tanx = sec^{2}x\)

Show us all your steps and we'll see if we can help.
 
I was using the sec and tan trig identities trying to simplify. Isn't sec[sup:2u0ncflj]2[/sup:2u0ncflj]x / tan x = sec x /sin x? Anyways, if I back up to before messing with trig identities I have:

y'/y = ln(x) sec[sup:2u0ncflj]2[/sup:2u0ncflj](x) / tan x + ln(tan(x))/ x

,from the product rule. Does that look right? If so, is there any simplification I can do or is that a reasonable answer?

I really appreciate your help. This Calc II class is really stretching my academic abilities and I'm trying to get as good of a grasp on all the homework as I can.
 
Ooh. Yes that looks fine. It isn't necessary to go through that much trouble in simplifying your expression but do as you see fit.
 
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