
#1
Dec903, 02:30 AM

P: 43

if y=x^TAN(x)
is y'=TAN(x)·x^(TAN(x)  1)·(SEC(x))^2 i got the answer by using the chain rule & the power rule by letting u=TAN(X) and y=x^u. not sure if thats the right answer since when i graph the function of y and y' , they don't show any relation with each other. i think i have to use the chain rule twice if i want to use the power rule, since x^u does not equal to ux^(u1) (i think, not exactly sure). can someone plz help me out, thanx. 



#2
Dec903, 02:49 AM

P: 265

If
[tex]y(x)=x^{\tan(x)}[/tex] then I get the following result for the derivative: [tex]y'(x)=x^{\tan(x)1}\left(x\log(x)(\sec(x))^2+\tan(x)\right)[/tex] or, if you prefer, [tex]y'(x)=x^{\tan(x)}\left(\log(x)(\sec(x))^2+\frac{\tan(x)}{x}\right)[/tex] I am not quite sure what you mean with: [tex]\frac{\partial}{\partial x} x^u = u x^{u1} \frac{\partial u}{\partial x}[/tex]. Cheers, Freek Suyver. 



#3
Dec903, 03:25 AM

P: 321

My answer is the same as that of suyver.
The first few steps should be: [tex] y = x^{tan x}[/tex] [tex]ln y = (tan x)(ln x)[/tex] {take log on both sides} Then you can apply the chain rule to finish the rest of the question. 



#5
Dec1003, 01:37 AM

P: 661

We can do it as suyver has done actually he has done the problem with partial differentiation




#6
Dec1303, 07:26 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

In general, if one has y= f(x)^{g(x)}, in which both base and exponent are functions of x, one can make either of two mistakes:
1. Treat the exponent, g(x), as a constant and use the power rule y'= g(x)f(x)^{g(x)1} 2. Treat the base, f(x), as a constant and use the exponential rule y'= ln(f(x))f(x)^{g(x)} The interesting thing is that the correct derivative is the sum of these two mistakes! y'= g(x)f(x)^{g(x)1}+ ln(f(x))f(x)^{g(x)} as one can show by differentiating ln(y)= g(x)ln(f(x)). 



#7
Dec1703, 03:25 PM

P: 28

my teacher taught me to use LN rather than LOG
so i ended up with [x^tan(x)]*(tan(x)/x + sec^2(x)ln(x)) 



#8
Dec1703, 03:36 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

Since there is no good reason to use logarithm to base 10 in higher mathematics (it's used in arithmetic because it works nicely with base 10 numeration), most higher mathematics texts use "log" to mean natural logarithm.




#9
Dec1703, 03:55 PM

P: 28





#10
Dec1803, 01:48 AM

P: 265





#11
Dec1803, 05:44 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,881

hodeez: Yes, I was responding to you. Sorry, I should have quoted your post. 



#12
Dec1803, 05:52 AM

P: 265




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