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Differentiating arctan

  • Thread starter ACLerok
  • Start date
I used implicit differentiating.

y = arctan(1/x)

is equivalent to

tan(y) = 1/x

but the second one is easier to differentiate. So I did that and solved for dy/dx to get

dy/dx = -cos^2(y)/x^2

Since tan(y) = 1/x, we can make a right triangle. Assign one angle to be y, so therefore the side opposite y will be equal to 1 and the adjacent side will be equal to x. You understand how to make this triangle?

By the pythagorean theorem, the hypotenuse of the triangle will be sqrt(x^2 + 1). So cos(y) is equal to x/sqrt(x^2 + 1), and therefore cos^2(y) = x^2/(x^2 + 1). Substitute this into the expression for dy/dx to get

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

which simplifies to

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

If you're still having trouble, you're going to need to be more specific about what's giving you difficulty.

cookiemonster
 
195
0
i dont know how you differentiated this: dy/dx = -(x^2/(x^2 + 1))(1/x^2)
and got: dy/dx = -1/(x^2 + 1)
 
195
0
haha nevermind i got it. Thanks for all your help!
 

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