# Need help on y=x^tan(x)

1. Dec 9, 2003

### Bailey

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^(u-1) (i think, not exactly sure).

can someone plz help me out, thanx.

2. Dec 9, 2003

### suyver

If

$$y(x)=x^{\tan(x)}$$

then I get the following result for the derivative:

$$y'(x)=x^{\tan(x)-1}\left(x\log(x)(\sec(x))^2+\tan(x)\right)$$

or, if you prefer,

$$y'(x)=x^{\tan(x)}\left(\log(x)(\sec(x))^2+\frac{\tan(x)}{x}\right)$$

I am not quite sure what you mean with:
but if you are talking about the derivative, then the statement is true:

$$\frac{\partial}{\partial x} x^u = u x^{u-1} \frac{\partial u}{\partial x}$$.

Cheers,
Freek Suyver.

3. Dec 9, 2003

### KLscilevothma

My answer is the same as that of suyver.

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:

$$y = x^{tan x}$$

$$ln y = (tan x)(ln x)$$ {take log on both sides}

Then you can apply the chain rule to finish the rest of the question.

4. Dec 9, 2003

### Bailey

thanx.

5. Dec 10, 2003

### himanshu121

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

6. Dec 13, 2003

### HallsofIvy

Staff Emeritus
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)).

Last edited: Dec 13, 2003
7. Dec 17, 2003

### hodeez

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. Dec 17, 2003

### HallsofIvy

Staff Emeritus
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. Dec 17, 2003

### hodeez

are you responding to me? if so, is my equation the equivalent to suyvers?

10. Dec 18, 2003

### suyver

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. Dec 18, 2003

### HallsofIvy

Staff Emeritus
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. Dec 18, 2003

### suyver

"Benifits" of a Dutch highschool 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 eachother and agree with the final result!