Let f be a differentiable function such that f(9) = 18 and f'(9) = 7. If g is the function defined by g(x) = f(x)/√x, what is the value of g'(9)?

Question

Let f be a differentiable function such that f(9) = 18 and f'(9) = 7. If g is the function defined by g(x) = f(x)/√x, what is the value of g'(9)?

A) 2

B) 7/3

C) 8/3

D) 42

Answer 1

what I know, if g(x)=f(x)/h(x) (here h(x) is square root x)

then g'(x)=[h(x)*f'(x)-h'(x)*f(x)]/(h(x))²

in words, derivative g=[h times derivative of f minus derivatve of h times f] divided by h squared

since h(x)=x^(1/2) g'(x)=[x^(1/2)*f'(x)-(1/2)*x^(-1/2)*f(x)]/x

Let f be a differentiable function such that f(9) = 18 and f'(9) = 7. If g is the function defined by g(x) = f(x)/√x, what is the value of g'(9)?

A) 2

B) 7/3

C) 8/3

D) 42

g'(9)=[3*7-(1/2)*(1/3)*18]/9=2

g'(9)=[3*f'(9)-(1/2)*(1/3)*f(9]/9=2

Answer 2

Good question. Get g'(x) by using the quotient rule. g'(x) = (√x * f'(x) - f(x) * 0.5/√x) / x

Next plug in 9.

g'(9) = (√9 * f'(9) - f(9) * 0.5/√9) / 9

g'(9) = (3 * 7 - 18 * 0.5/3) / 9 = (21 - 3)/9 = 2