If f(x) = 5/x^2, and g(x) =f^-1(x), find the slope of the curve g(x) = f^-1(x) at the point (5

If f(x) = 5/x^2, and g(x) =f^-1(x), find the slope of the curve g(x) = f ^-1(x) at the point (5, 1).

There is a theorem that says g'(y) = 1/f '(x) when g(x) is the inverse of f(x)

f(x)= 5/x^2= 5x^-2
so (1,5) is a point on f(x) and (5,1) is a point on g(x). The derivatives will be reciprocals.

f '(x)= -10x^-3

f '(1)= -10

g'(5)= -1/10

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Or finding it directly.

g(x)= (5/x)^(1/2)= (x/5)^(-1/2)

g'(x)= (-1/2)(x/5)^(-3/2) * (1/5) = (-1/10)(x/5)^(-3/2)

g'(5)= -1/10

Hoping this helps!