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

[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 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.

 

[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:

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:

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

Cheers,
Freek Suyver.

 

[KLscilevothma]

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:

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.

 

[Bailey]

thanx.

 

[himanshu121]

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

 

[HallsofIvy]

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)).

 

[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))

 

[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.

 

[hodeez]

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?

 

[suyver]

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.

 

[HallsofIvy]

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.

 

[suyver]

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!