Question

how do you get the derivative of ln(tanx)? because i got 1/tanx*sec^2x but it's wrong?

Answer 1

well isn't tan = sin/cos

and I think you need to look at the log laws

so you are looking at

\[ \ln \frac{\sin(x)}{\cos(x)} = \ln(\sin(x)) - \ln(\cos(x))\]

now you can take the derivative

Answer 2

or you use the chain rule

derivative = (1/ tanx ) * sec^2x

Answer 3

= 1 cos x
---- * ---- = 1 / cosx sinx
cos^2 sinx


is one way of writing it

Answer 4

the derivative of that just take the inside derivative which is 1+tan^2x
then the derivative of the outer function which is 1/tanx
so \(\Large {\frac{1+tan^2x}{tanx}}=\normalsize cotx+tanx\)

Answer 5

derivative of tanx is (secx)^2 which is also 1+(tanx)^2

Answer 6

your answer is ok. There are a few ways to write it.

Answer 7

thanks guys but the answer is 2/sin2x , because i have a maths book with all the answers written in the back of the book, just not the methods... i literally have no idea to go from the answer i got to that answer... help pls

Answer 8

yes - many ways

Answer 9

sin(2x) = 2 sin(x) cos(x)

Answer 10

you can also use tanx=sinx/cosx like @campbell_st did
Ate any rate just be aware that there are many ways to right trig functions
using identities like @phi said

Answer 11

and 2/(2 sin x cos x) = 1/(cos x sin x)

Answer 12

yes 2 / sin2x = 2 / 2sinx cosx = 1 / sinx cosx

Answer 13

Here is a graph of (1/tan x) * sec^2(x) and 2/sin(2x)

Answer 14

@phi may i ask what graphing web you used?

Answer 15

geogebra. It's free. google for it.

Answer 16

oh ok! thanks

Answer 17

ok so now i wrote it as 1+tan^2x/tanx ... now what?

Answer 18

you can leave is sec^2x/tanx or put in any other different form you like
there are all correct

Answer 19

i just want to know how to get to 2/sin2x ,i'm really confused lol sorry i'm not that good at maths and we've just started differentiating 2 days ago

Answer 20

\[\frac{\sec^2(x)}{\tan(x)} =\frac{1}{\cos^2(x)} \cdot \frac{\cos(x)}{\sin(x)}\]
But recall 2sin(x)cos(x)=sin(2x)
so sin(x)cos(x)=sin(2x)/2

Answer 21

@freckles where did you get 1/cos^\[1/\cos^2x . cosx/sinx ?\]

Answer 22

sec^2(x)=1/cos^2(x)

Answer 23

1/tan(x)=cos(x)/sin(x)

Answer 24

@freckles how is sec^2x equal to 1/cos^2x? and how does tan become 1/tan(x) and how does 1/tanx become cosx/sinx?? sorry it doesn't make sense to me... and when u multiply cos^2x by sinx, where does the 2 go?

Answer 25

\[\frac{\sec^2(x) \cdot 1}{1 \cdot \tan(x)}=\frac{\sec^2(x)}{1} \cdot \frac{1}{\tan(x)}\]

Answer 26

\[\sec(x)=\frac{1}{\cos(x)} \\ \text{ squaring both sides gives } \\ (\sec(x))^2=(\frac{1}{\cos(x)})^2 \\ (\sec(x))^2=\frac{1^2}{(\cos(x))^2} \\ (\sec(x))^2=\frac{1}{(\cos(x))^2} \\ \sec^2(x)=\frac{1}{\cos^2(x)}\]

Answer 27

\[\frac{1}{\tan(x)}=\cot(x)=\frac{\cos(x)}{\sin(x)}\]

Answer 28

\[\frac{\sec^2(x)}{\tan(x)}=\sec^2(x)\cot(x)=\frac{1}{\cos^2(x)} \frac{\cos(x)}{\sin(x)} \\ =\frac{\cos(x)}{\cos(x)\cos(x)\sin(x)} \\ =\frac{ \cancel {\cos(x)}}{\cancel{\cos(x)} \cos(x)\sin(x)}\]

Answer 29

\[=\frac{1}{\frac{1}{2} 2 \sin(x)\cos(x)}\]
You should be able to finish this...

Answer 30

given sin(2x)=2sin(x)cos(x)

Answer 31

I take it you didn't take trigonometry before calculus ?

Answer 32

This means you are going to have learn trig on your own (or with the help of OpenStudy) without the instructor going over the trig like stuff.

Answer 33

\[\sin(\theta)=\frac{y}{r} \\ \cos(\theta)=\frac{x}{r} \\ \tan(\theta)=\frac{y}{x}=\frac{\frac{y}{r}}{\frac{x}{r}}=\frac{\sin(\theta)}{\cos(\theta)} \\ \csc(\theta)=\frac{r}{y}=\frac{1}{\frac{y}{r}}=\frac{1}{\sin(\theta)} \\ \sec(\theta)=\frac{r}{x}=\frac{1}{\frac{x}{r}}=\frac{1}{\cos(\theta)} \\ \cot(\theta)=\frac{x}{y}=\frac{\frac{x}{r}}{\frac{y}{r}}=\frac{\cos(\theta)}{\sin(\theta)}\]