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

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Answer 1 · 2017-07-12T00:53:03

I got

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

I'll assume you don't know that #d/(dx)[arc cot(u(x))] = -1/(1 + u^2)(du)/(dx)#.

Instead, I'll rewrite this as

#coty = x/5#

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

Therefore:

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

And since we defined #y = arc cot(x/5)#, we can use the identity

#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 #arc cosu#, you would still get

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

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