# quick double check on trig function derivatives

• Mar 8th 2007, 06:39 PM
UMStudent
quick double check on trig function derivatives
Ok so I know the derivative of tanx = sec^2(x) so then would the tangent of tan (x^2) = sec^2(x^2)? or sec^4(x). And then the derivative of that would be ?

I know the derivative of sec(x) = sec(x) tan(x) so sec^2(x) would have derivative sec^2(x)tan^2(x)?

I'm doing derivative of a derivative. I'm normally not so in need of help with math but I ended up in the hospital for a week and kinda feel behind. Thanks for all the help lately on here.

Oh, and the original problem was tan^-1(x^2) so then 1/tan(x^2). Just wondering from there so I can double check my answers.
• Mar 8th 2007, 06:43 PM
Jhevon
Quote:

Originally Posted by UMStudent
Ok so I know the derivative of tanx = sec^2(x) so then would the tangent of tan (x^2) = sec^2(x^2)? or sec^4(x). And then the derivative of that would be ?

I know the derivative of sec(x) = sec(x) tan(x) so sec^2(x) would have derivative sec^2(x)tan^2(x)?

I'm doing derivative of a derivative. I'm normally not so in need of help with math but I ended up in the hospital for a week and kinda feel behind. Thanks for all the help lately on here.

Oh, and the original problem was tan^-1(x^2) so then 1/tan(x^2). Just wondering from there so I can double check my answers.

Ok, so in the case of something like tan(x^2), we would use the chain rule, so the derivative would be 2x*sec^2(x^2).

recall that the chain rule says d/dx (f(g(x))) = f '(g(x))*g'(x). what do you think the others will be?
• Mar 8th 2007, 06:58 PM
Aryth
Quote:

Originally Posted by UMStudent
Ok so I know the derivative of tanx = sec^2(x) so then would the tangent of tan (x^2) = sec^2(x^2)? or sec^4(x). And then the derivative of that would be ?

I know the derivative of sec(x) = sec(x) tan(x) so sec^2(x) would have derivative sec^2(x)tan^2(x)?

I'm doing derivative of a derivative. I'm normally not so in need of help with math but I ended up in the hospital for a week and kinda feel behind. Thanks for all the help lately on here.

Oh, and the original problem was tan^-1(x^2) so then 1/tan(x^2). Just wondering from there so I can double check my answers.

Well, let's take tan(x^2) and find the derivative of it to see if you are correct:

d/dx tan u = sec^2 u du/dx

Now we plug in x^2 for u

d/dx tan(x^2) = sec^2(x^2) d(x^2)/dx

d/dx tan(x^2) = 2x*sec^2(x^2)

For sec^2(x), recall that it is the same as (sec(x))^2 so that:

d/dx [u]^2 = 2[u] du/dx

Now we plug in sec(x) into u

d/dx sec^2(x) = 2sec(x) d(sec(x))/dx

d/dx sec^2(x) = 2sec^2(x)tan(x)

Now, for 1/tan(x^2) we use the quotient rule, just because it's easier for me:

d/dx 1/tan(x^2) = (1)'(tan(x^2)) - (1)(tan(x^2)' / tan^2(x^2)

d/dx 1/tan(x^2) = (0)(tan(x^2)) - (1)(2xsec^2(x^2))/tan^2(x^2)

d/dx 1/tan(x^2) = -2x*sec^2(x^2) / tan^2(x^2)

• Mar 8th 2007, 07:09 PM
UMStudent
Ok so I'm guessing we need to use the product rule on this one. So...

2x sec^2(x^2)
2x d/dx sec^2(x^2) + sec^2(x^2) d/dx (2x)
2x * 2x sec^2(x^2)tan^2(x^2) + sec^2(x^2) *2
4xsec^2(x^2)tan^2(x^2) + 2 sec^2(x^2)

Close at all?

I'm sure I'm wrong because my book gives limited information. If I had some basic rules on derivatives of trig functions it would help like what to do if I have a number x sin ^a(b) or x sin^a(b^c) or just sin^2 x. Do the powers stay whats a constant or do I use other derivative rules the same way I use on say x^3 = 3x^2. There is nothing in the book so I don't know.
• Mar 8th 2007, 07:20 PM
Aryth
Quote:

Originally Posted by UMStudent
Ok so I'm guessing we need to use the product rule on this one. So...

2x sec^2(x^2)
2x d/dx sec^2(x^2) + sec^2(x^2) d/dx (2x)
2x * 2x sec^2(x^2)tan^2(x^2) + sec^2(x^2) *2
4xsec^2(x^2)tan^2(x^2) + 2 sec^2(x^2)

Close at all?

I'm sure I'm wrong because my book gives limited information. If I had some basic rules on derivatives of trig functions it would help like what to do if I have a number x sin ^a(b) or x sin^a(b^c) or just sin^2 x. Do the powers stay whats a constant or do I use other derivative rules the same way I use on say x^3 = 3x^2. There is nothing in the book so I don't know.

Yes, in fact you would use the product rule. So it gives you this:

d/dx 2xsec^2(x^2) = (d(2x)/dx)(sec^2(x^2)) + (2x)(d(sec^2(x^2)/dx)

For d/dx (sec(x^2))^2 = 2[u]*d[u]/dx

For d/dx sec(u) = sec(u)tan(u) du/dx
d/dx sec(x^2) = 2x*sec(x^2)tan(x^2)

d/dx (sec(x^2))^2 = 2sec(x^2)*2x sec(x^2)tan(x^2)
=4x*sec^2(x^2)tan(x^2)

d/dx 2x*sec^2(x^2) = 2*sec^2(x^2) + 2x(4x*sec^2(x^2)tan(x^2))

d/dx 2x*sec^2(x^2) = 2*sec^2(x^2) + 8x^2*sec^2(x^2)tan(x^2)

You can factor out a 2*sec^2(x^2) if you wish to simplify it a bit:

d/dx 2x*sec^2(x^2) = 2*sec^2(x^2)( 1 + 4x^2*tan(x^2)

• Mar 8th 2007, 07:43 PM
UMStudent
Ok so I'm trying to understand this and am starting to get it. So like the derivative of sin^2(x^2) = 2x cos^2(x^2)? Just trying to understand some simple rules like ^2(x^2) is a constant and the derivative of x^2 = 2x so we move that to the front of the statement? Would that be a basic rule I could use so? Or another one the derivative of sin^2 x is cos^2 x so all that changes is the trig function and the power is a constant?

And if I understand the above correctly then:

tan^-1(x^2) = 1/tan(x^2) and the derivative of that is
(using the quotient rule)
1 * d/dx tan(x^2) - tan(x^2) * d/dx 1 / tan^2(x^2)
d/dx of 1 = 0 so the left side then pretty much cancles out we are left with
1 * d/dx tan(x^2) / tan^2(x^2)
2x sec^2(x^2)/ tan^2(x^2)

and then i'm lost because i don't know what to do if tan is squared
d/dx of tan(x) = sec^2(x) so would tan^2(x) = sec^4(x)??

if correct then I could continue
2x sec^2(x^2)/2x sec^4(x^2)? and that would be the derivative of 1/tan(x^2)?

Then I have to take the derivative again. But lets make sure i'm right so far, and I understand some of this basic rules.
• Mar 8th 2007, 09:06 PM
UMStudent
Ok I am finally understanding what is going on now I found some other rules in the book I had been previously overlooking so I got the derivative of tan^1(x^2) = -2xsec^2(x^2)/tan^2(x^2) and I understand how I got that but now to take the derivative of that would I have to use the product rule for the top and then also use the quotient rule. Or actually I'm guessing thats way wrong and its just the quotient rule I use again correct? Well actually looking at the problem now I guess I will be using the quotient rule but then using the product rule inside that to find the derivative of like -2x sec^2(x^2) right?
• Mar 9th 2007, 08:15 AM
Aryth
Quote:

Originally Posted by UMStudent
Ok I am finally understanding what is going on now I found some other rules in the book I had been previously overlooking so I got the derivative of tan^1(x^2) = -2xsec^2(x^2)/tan^2(x^2) and I understand how I got that but now to take the derivative of that would I have to use the product rule for the top and then also use the quotient rule. Or actually I'm guessing thats way wrong and its just the quotient rule I use again correct? Well actually looking at the problem now I guess I will be using the quotient rule but then using the product rule inside that to find the derivative of like -2x sec^2(x^2) right?

Yes, to find the derivative of -2x*sec^2(x^2)/tan^2(x^2) you would have to use the quotient rule:

tan^2(x^2)(d(-2x*sec^2(x^2)/dx) - (2x*sec^2(x^2))*(d(tan^2(x^2)/dx)/(tan^2(x^2))^2

To get the derivative of -2x*sec^2(x^2) you have to use the product rule:

(d(-2x)/dx)(sec^2(x^2)) + (-2x)(d(sec^2(x^2))/dx)

(-2)(sec^2(x^2)) + (-2x)(2sec(x^2)*(2xsec(x^2)tan(x^2))

-2*sec^2(x^2) -8x*sec^2(x^2)tan(x^2)

You can factor out a -2*sec^2(x^2) to get:

d/dx 2x*sec^2(x^2) = -2*sec^2 (1 + 4x*tan(x^2))

There is your derivative for that, and now we take the derivative of tan^2(x^2):

d/dx [u]^2 = 2[u] d[u]/dx

d/dx (tan(x^2))^2 = 2*tan(x^2) (d[tan(x^2)]/dx)

d/dx tan(x^2) = sec^2(x^2)*2x

d/dx tan^2(x^2) = 2*tan(x^2)*2x*sec^2(x^2)

d/dx tan^2(x^2) = 4x*tan(x^2)sec^2(x^2)

Now we plug them both into the quotient rule:

tan^2(x^2)(d(-2x*sec^2(x^2)/dx) - (2x*sec^2(x^2))*(d(tan^2(x^2)/dx)/(tan^2(x^2))^2

tan^2(x^2)(-2*sec^2(x^2) (1 + 4x*tan(x^2)) - (2x*sec^2(x^2))*(4x*tan(x^2)sec^2(x^2))/(tan^2(x^2))^2

(tan^2(x^2)*-2*sec^2(x^2) + 4x*tan^2(x^2)sec^2(x^2)) - (8x*sec^4(x^2)tan(x^2)/ tan^4(x^2)

(2x*tan^2(x^2)sec^2(x^2)) - (8x*sec^4(x^2)tan(x^2)) / tan^4(x^2)

(2x*sec^2(x^2))/(tan^3(x^2)) - (8x*sec^4(x^2))/(tan^3(x^2))

1/(tan^3(x^2))*(2x*sec^2(x^2)-8x*sec^4(x^2))

And there's your derivative. If you notice, the more you get the derivative of this stuff, the more complicated it gets.