if f(x) and g(x) are inverses of each other
how does f'(x)=1/g'(X) ?
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.
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)
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.
To understand why f'(x) = 1/g'(x) if f(x) and g(x) are inverses of each other, we need to use the concept of the chain rule in calculus.
Let's start by considering the compositions of the two functions f(x) and g(x):
1. f(g(x)) = x (Since g(x) is an inverse of f(x))
2. g(f(x)) = x (Since f(x) is an inverse of g(x))
Now, let's find the derivatives of these compositions using the chain rule:
1. Differentiating f(g(x)) with respect to x:
Applying the chain rule, we have:
f'(g(x)) * g'(x) = 1
2. Differentiating g(f(x)) with respect to x:
Applying the chain rule, we have:
g'(f(x)) * f'(x) = 1
Since we know that f(g(x)) = x and g(f(x)) = x, we can equate the two equations:
f'(g(x)) * g'(x) = 1
g'(f(x)) * f'(x) = 1
Since f(x) and g(x) are inverses, g(x) = f^(-1)(x), so we can substitute g(x) with f^(-1)(x) in the first equation and f(x) with g^(-1)(x) in the second equation:
f'(f^(-1)(x)) * f^(-1)'(x) = 1
--> f^(-1)'(x) = 1 / f'(f^(-1)(x))
Similarly,
g'(g^(-1)(x)) * g^(-1)'(x) = 1
--> g^(-1)'(x) = 1 / g'(g^(-1)(x))
Since f^(-1)(x) = g(x) and g^(-1)(x) = f(x), we can rewrite the equations as:
f^(-1)'(x) = 1 / f'(g(x))
g^(-1)'(x) = 1 / g'(f(x))
Therefore, we can conclude that if f(x) and g(x) are inverses of each other, then f'(x) = 1 / g'(x) and g'(x) = 1 / f'(x).