Question

Consider the function

\(f(x) = \tan(x) := \tan(x) := \frac{sin(x)}{cos(x)}\)

Determine the derivative \(\frac{d}{dx}tan(x)\)

Answer 1

do you have to show or no, because this is one of the derivatives you should know

it's sec^2(x)

Answer 2

if you do it with sin(x)cos(x) you have to simply do product rule

Answer 3

hmm..

Answer 4

(sinx*d(cosx)/dx - cosx*d(sinx)/dx)/cos^2(x)

Answer 5

\[uv'+vu'=sin(x)(-1)(cos(x)^{-2}(-sinx)+cos(x)^{-1}*cos(x)\]

Answer 6

i think two different aspect right?

Answer 7

sorry I did it backwards
(
cosx*d(sinx)/dx- sinx*d(cosx)/dx)/cos^2x

Answer 8

(cos^2(x) + sin^2(x))/cos^2(x)

Answer 9

pythagorean theorem says the top is 1
so 1/cos^2(x) = sec^2(x)

Answer 10

@zzr0ck3r you use \[\frac{1}{\cos^{2x}}\] is it that what u wrote as a last one ?

Answer 11

@Outkast3r09 are you agree with the solution of zzrOck3r?

Answer 12

@zzr0ck3r @Outkast3r09 thank you guys..

Answer 13

check this out also. http://www.wolframalpha.com/input/?i=derivative+cos%28cos%28x%29%29 this site will help you for all calc

Answer 14

hit the show steps button. the walkthrough is not so great on tanx but any other should be good, and this site will be great next term in integral calc.

Answer 15

this site gives but not the steps right, i think it gives just the conclusion?

Answer 16

I know for a fact that is the answer, because my book has it as a main derivative , however if you didn't know what it was you can differentiate sin/cos and you'll get the answer still

Answer 17

I can give you this list if you'd like tunahan.. i'm suprised your book doesn't have it @tunahan

Answer 18

thanks @Outkast3r09 to be honest, i dont had time to learn math and i am quite bad at it, (this things sure somewhere in our books too) but in this days i must start to learn because of exam 18.of july