Master the Chain and Power Rules: Simplifying y=x^tan(x) and Solving for y

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Discussion Overview

The discussion revolves around finding the derivative of the function y = x^tan(x) using the chain rule and power rule. Participants explore different methods for differentiation, including the use of logarithms, and express uncertainty about their approaches and results.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant proposes that the derivative y' = TAN(x)·x^(TAN(x) - 1)·(SEC(x))^2 is obtained using the chain and power rules, but expresses uncertainty about the correctness of this result due to a lack of correlation between the graphs of y and y'.
  • Another participant provides a different expression for the derivative, y' = x^{tan(x)-1}(x log(x)(sec(x))^2 + tan(x)), and clarifies that the derivative requires careful application of the chain rule and power rule, particularly noting that u = tan(x) should not be treated as a constant.
  • A third participant agrees with the second and suggests using the natural logarithm to differentiate, indicating that taking the logarithm of both sides can simplify the differentiation process.
  • There is a discussion about the use of logarithms, with some participants noting the preference for natural logarithm (ln) over logarithm base 10 (log), and clarifying their definitions of logarithmic notation.
  • One participant mentions that the correct derivative can be seen as a combination of two common mistakes in differentiation involving functions of x in both the base and exponent.

Areas of Agreement / Disagreement

Participants express differing views on the correct method for differentiating the function, with some advocating for the use of logarithms while others question the treatment of tan(x) as a constant. There is no consensus on the best approach or the correctness of the various proposed derivatives.

Contextual Notes

Participants highlight the importance of understanding the roles of the base and exponent in differentiation, and the discussion reflects varying educational backgrounds regarding logarithmic notation and differentiation techniques.

Bailey
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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 that's 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^(u-1) (i think, not exactly sure).

can someone please help me out, thanx.
 
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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:
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^(u-1) (i think, not exactly sure).
but if you are talking about the derivative, then the statement is true:

[tex]\frac{\partial}{\partial x} x^u = u x^{u-1} \frac{\partial u}{\partial x}[/tex].

Cheers,
Freek Suyver.
 
My answer is the same as that of suyver.

i got the answer by using the chain rule & the power rule by letting u=TAN(X) and y=x^u.
It isn't correct. You cannot let u = (tan x) and then treat it as a constant, because y is a function of x and (tan x) isn't a constant. So we shouldn't apply the power rule directly. Instead, we should first take ln(natural log) on both sides, and then do differentiation.
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.
 
thanx.
 
We can do it as suyver has done actually he has done the problem with partial differentiation
It isn't correct. You cannot let u = (tan x) and then treat it as a constant
 
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)).
 
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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))
 
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.
 
Originally posted by HallsofIvy
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.

are you responding to me? if so, is my equation the equivalent to suyvers?
 
  • #10
Originally posted by hodeez
are you responding to me? if so, is my equation the equivalent to suyvers?

Yes, your equation is equivalent to mine. I always write log(x) for the e-based logarithm and 10log(x) for the 10-based logarithm.
 
  • #11
originally posted by suyver
I always write log(x) for the e-based logarithm and 10log(x) for the 10-based logarithm.

Now that's really strange! Most people use log10(x)!

hodeez: Yes, I was responding to you. Sorry, I should have quoted your post.
 
  • #12
Originally posted by HallsofIvy
Now that's really strange! Most people use log10(x)!

"Benifits" of a Dutch high school education: nlog(x) is the n-based logarithm; log(x) is the 10-based and ln(x) is e-based. At university the e-based logarithm becomes log(x) and people stop using ln(x), but still nlog(x) remains the n-based logarithm. Oh well, just as long as we understand each other and agree with the final result! :smile:
 

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