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 .....
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.
3 answers
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?
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?
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
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