# How do you differentiate #y=arc cot(x/5)#?

##### 1 Answer

Jul 12, 2017

I got

#(dy)/(dx) = -1/5 (1/(1 + (x/5)^2))#

I'll assume you don't know that

Instead, I'll rewrite this as

#coty = x/5#

Then, by implicit differentiation:

#-csc^2y (dy)/(dx) = 1/5#

Therefore:

#(dy)/(dx) = -1/(5csc^2y)#

And since we defined

#csc^2y = 1 + cot^2y#

to get

#color(blue)((dy)/(dx)) = -1/5 1/(1 + cot^2y)#

#= color(blue)(-1/5 (1/(1 + (x/5)^2)))#

If you did it using the actual derivative of

#(dy)/(dx) = -1/(1 + underbrace((x/5)^2)_(u^2)) cdot d/(dx)[x/5]#

#= -1/5 (1/(1 + (x/5)^2))#