if f(x) and g(x) are inverses of each other

Question

how does f'(x)=1/g'(X) ?

Answer 1

If f(x) and g(x) are inverses of each other, it means that

f(g(x)) = x (1)

g(f(x)) = x (2)

Taking the derivative of equation (1) with respect to x:

f'(g(x)) * g'(x) = 1

f'(g(x)) = 1/g'(x)

Taking the derivative of equation (2) with respect to x:

g'(f(x)) * f'(x) = 1

f'(x) = 1/g'(f(x))

Let y = g(x), then equation (1) becomes:

f(y) = x

Differentiating with respect to y:

f'(y) * dy/dx = 1

dy/dx = 1/f'(y)

Substituting y = g(x):

dy/dx = 1/f'(g(x))

Using the equation derived from equation (2):

f'(x) = 1/g'(f(x))

Substituting f(x) = y:

f'(y) = 1/g'(x)

Substituting back into the equation for dy/dx:

dy/dx = 1/f'(y) = 1/g'(x)

Therefore, f'(x) = 1/g'(x) if f(x) and g(x) are inverses of each other.

Answer 2

actually, if g(x) = f^-1(x) then
if f(a) = b then
g'(b) = 1/f'(a)

It is not true in general that g'(x) = 1/f'(x)
consider
f(x) = x^3
g(x) = ∛x
f'(x) = 3x^2
g'(x) = 1/(3∛x^2) ≠ 1/(3x^2)

But note that since f(1) = g(1), we do have g'(1) = 1/f'(1)
But since f(2) = 8
g'(8) = 1/(3∛64) = 1/12 = 1/f'(2)

Answer 3

Thank you for the clarification. Yes, you are correct that the correct statement is:

If g(x) = f^-1(x), then for any b = '(b) = 1/f'(a).

This implies that g'(x) = 1/f'(g(x)) only at the point g(x) = a, where f(a) = x.

Your example with f(x) = x^3 and g(x) = ∛x is a great illustration of this concept.