2a)
I will assume you mean f(x) = 3/(x+1)
I would write it f(x) = 3(x+1)^-1
then f ' (x) = -3(x+1)^-2
= -3/(x+1)^2
at P(2,1), slope = -3/(3)^2 = -3/9 = -1/3
h(x) = 2/√(x+5)
= 2(x+5)^(-1/2)
h ' (x) = -1(x+5)^(-3/2)
= -1/( √(x+5) )^3
you finish it
Hellp I have a calculus test on Monday
2. Calculate the slope of the tangent to the given function at the given point or value of x
a. f(x)=3/x+1,P(2,1)
b.h(x)=2/squareroot x + 5, P(4,2/3)
****Full solutions to please because I do not know what I'm doing and so confused with calculus
2 answers
how you get f ' (x) = -3(x+1)^-2 what numbers do i multiply with
3 answers