# Differentiating arctan

#### cookiemonster

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

#### ACLerok

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)

#### ACLerok

haha nevermind i got it. Thanks for all your help!

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving