Let #f(x) = x^2 - 4x - 5#, x > 2, how do you find the value of #(df^-1)/dx# (or the derivative of the inverse of f(x)), at the point x = 0 = f(5)?
1 Answer
Explanation:
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G(x) = cos(x/2) - 3x f(x) = 2x + 5 g(f(x)) = cos(f(x)/2) - 3f(x) = cos(2x+5)/2 - 3(2x+5) = cos(2x+5)/2 -6x-15.
we have #f(x) = x^2 - 4x - 5# and you want the derivative of its inverse
so if
to that end we can say that
and using the chain rule:
so #g'(y) =(f^(-1)(x))^prime = = 1/(f'(x))#
#=1/( 2x - 4)#
there's poss another way to show this, which is a bit of a desecration of the Liebnitz notation: ie #dx/dy = 1/ (dy/dx)# , really not sure that means very much though as #dy/dx# is not a fraction with a numerator and denominator