Question

Let f(x)=5+3/x

a) Find f′(x)

b)Find the slope of the tangent line at x=1. Slope = f′(1)=?

Answer 1

solution

f(x)=5+3/x

a) Find f′(x)

d(f(x)/dx = f'(x) = d(5)/dx + d(3/x)/dx

derivative of constant is zero = d(5)/dx =0

d(x^n) = nx^n-1 here n =-1

d(1/x) = d(x^-1) = -1x^(-1-1) = -1x^-2 = -1/x^2

d(1/x)/dx = -1/x^2

= d(5)/dx + 3d(1/x)/dx = 0 +3(-1/x^2) = -3/x^2

d(f(x)/dx = f'(x) = d(5)/dx + d(3/x)/dx = -3/x^2

derivative of a function f(x) is slope = d(f(x)/dx = f'(x) = -3/x^2

b)Find the slope of the tangent line at x=1. Slope = f′(1)=?

derivative of a function f(x) is slope = d(f(x)/dx = f'(x) = -3/x^2

f'(1) put x =1

f'(1) = -3/(1^2) =-3

Slope = f′(1)= -3