Since [tex]f^{-1}[/tex] is a reflection of [tex]f[/tex] about the line y=x, we can say that the derivative (slope of tangent at any point) of [tex]f^{-1}[/tex] is the reciprocal of the derivative of [tex]f[/tex].(adsbygoogle = window.adsbygoogle || []).push({});

Now suppose [tex]y=x^3+2x^2+4[/tex]

so that [tex]\frac{dy}{dx} = 3x^2+4x[/tex]

But then why am I wrong in saying that [tex] ( f^{-1} )'(x) = \frac{1}{ \frac{dy}{dx} } = \frac{1}{3x^2+4x}?[/tex]

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Relationship between the derivative of f and the derivative of f inverse

**Physics Forums | Science Articles, Homework Help, Discussion**