Question

differentiate g(x) = tan x - cot x, differentiate

Answer 1

2(sec^2x tan^2x) - 2(csc^2x cot^2x) by the chain rule.

Answer 2

the answer is g'x = 4 csc^2(2x)

Answer 3

woops the question was g(x) = tan x - cot x, differentiate

Answer 4

Yah this one is a little tricky kels, you either need to REMEMBER the derivative of tanx and cotx (which I would recommend :) ) Or you can convert everything to sines and cosines to get the job done.

Answer 5

g'(x) = sec^2x +cosec^2x= (sin^2x+cos^2x) /((sin^2x)(cos^2x))
=4/4(((sin^2x)(cos^2x))=4(cosec2x)^2

Answer 6

i got stuck at sec^2x +cosec^2x.. i dont think i'll ever understand this stuff

Answer 7

I believe it is easier if you convert it first, make it simpler then derivate.\[g(x)=\tan x-\cot x=\frac{\sin x}{\cos x}-\frac{\cos x}{\sin x}=-\frac{\cos^2 x -\sin^2 x}{\cos x \sin x}=-\frac{2\cos (2x)}{\sin (2x)}\]Derivating it you get:\[g'(x)=-2\left[\frac{-2\sin(2x)}{\sin(2x)}+\frac{-2\cos^2(2x)}{\sin^2(2x)}\right]=4+4\cot^2(2x)\]

Answer 8

im stuck at how cos2x−sin2x/cosxsinx = −2cos(2x)/sin(2x)..

Answer 9

Double Angle Formula for Cosine on top, for Sine on bottom :O