Question

find an equation for the line tangent to the graph of y=arccot(x) at x=1.

I found that slope is -1/(1+x^2). I don't know how to evaluate arccot(1) to find y value.

Answer 1

arccot(1) means give me the value or the angle so that cot(that value) = 1
we know that tan(π/4) = 1 , (tan 45° = 1)
and since cotØ = 1/tanØ

arccot(1) = π/4 , thus y = π/4

so now you have the slope and the point (1,π/4)
carry on .....

Answer 2

Thank you very much! I get it now. For another question:

y'=lim h->0 ((e^x+h)-e^x)/h. So I simplified this and got (e^x)* lim h->0 ((e^h)-1)/h. But I don't understand why and how ((e^h)-1)/h equals 1. Explain?

Answer 3

if y = e^x
the derivative of y in terms of fundamental concepts or sometimes called by "first principles"
lim ( e^(x+h) - e^x)/h as h --->0 , (notice the proper placement of brackets)
so you are simply given the dy/dx of e^x
thus y = e^x