Question

What is the derivative of y = x/tanx ?

Answer 1

I'm not sure if it is correct
y = x/tanx
\[\frac{dy}{dx} = \frac{tan x \frac{d}{dx} x - x\frac{d}{dx}tanx}{tan^2x} = \frac{tan x - xsec^2x}{tan^2x}\]
Perhaps, you need to simplify it...

Answer 2

it says the answer is y'= cot(x) -x*csc^2x.....i just wanted to know how i can get the original function to this.

Answer 3

Just simplify it ..... First of all, do you understand the above steps?

Answer 4

not really

Answer 5

how did u get tanx - xsec^2x

Answer 6

\[\frac{d}{dx} x = 1\]So, \[tan x \frac{d}{dx} x = tanx (1) = tanx\]
\[\frac{d}{dx}tanx = sec^2x\]So, \[x\frac{d}{dx}tanx=xsec^2x\]

From the above, you'll get
\[tan x \frac{d}{dx} x - x\frac{d}{dx}tanx = tanx - xsec^2x\]

Answer 7

that makes sense but how would it become y' = cot(x) - x*csc^2x ?

Answer 8

From the first comment, we've got
\[y' = \frac{tan x - xsec^2x}{tan^2x} \]\[y' = \frac{tan x - xsec^2x}{tan^2x} = \frac{tan x }{tan^2x} - \frac{xsec^2x}{tan^2x}\] Can you simplify it?

Note: \[(i) \frac{1}{tanx} = cotx\]\[(ii)secx =\frac{1}{cosx} \]\[(iii) \frac{1}{tanx} = \frac{cosx}{sinx}\]

Answer 9

if 1/tanx = cot x then what does tan^2x equal to ?

Answer 10

nevermind i figured it out ! thank you so much !!!! :)

Answer 11

Sorry that I didn't notice your reply :(
Glad that you've figured it though !!