Suppose h(x)=square root of f(x) and the equation of the tangent line to f(x) at x=1 is y=4+3(x-1) Find h’(1)

Answer 1

h'(x) = f(x)^-1/2·f'(x)/2; h'(1) = f(1)^-1/2f'(1)/2. Now from equation of tangent line f(1) = 4 and f'(1) = 3.

So h'(1) = 4^-1/2·3/2 = 3/4