Question #fc13e

2 Answers
Jim H
Oct 23, 2016

Let #y = arccot x#

Then #x = cot y#.

Differentiate implicitly to get

#1 = -csc^2 y * dy/dx#

So that #dy/dx = -1/csc^2 y#

Recall from trigonometry that #1+cot^2y = csc^2 y# so

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

Finally replace #cscy# with #x# (from the second line of the explanation.)

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

sente
Oct 23, 2016

Explanation:

We'll use implicit differentiation:

Let #y = "arccot"(x)#

#=> cot(y) = x#

#=> d/dxcot(y) = d/dxx#

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

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

Now let's see what #csc^2(y)# is in terms of #x#. Draw a right triangle with an angle #y# such that #cot(y) = x# (i.e. #y= "arccot"(x)#):

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Setting the legs so that #cot(y) = x#, we can find the length of the hypotenuse as #sqrt(1+x^2)# using the Pythagorean theorem. Now we can calculate #csc^2(y)#.

#csc^2(y) = (csc(y))^2 = (sqrt(1+x^2)/1)^2 = 1+x^2#

Substituting this in, we arrive at our answer:

#("arccot"(x))' = dy/dx = -1/(1+x^2)#

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