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

[Juwane]

Since $$f^{-1}$$ is a reflection of $$f$$ about the line y=x, we can say that the derivative (slope of tangent at any point) of $$f^{-1}$$ is the reciprocal of the derivative of $$f$$.

Now suppose $$y=x^3+2x^2+4$$

so that $$\frac{dy}{dx} = 3x^2+4x$$

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

 

[JSuarez]

Because the inverse's derivative is a function of y, not x.

 

[Juwane]

Because the inverse's derivative is a function of y, not x.

But if we look at the curve of the inverse, we'll see that to find the tangent at any point on the curve, we'll have to replace x with with that point. I mean, the tangent is still dy/dx, because the axes are the same: y and x.

 

[elibj123]

The tangent line will be the same, but its slope will change, because now you calculate the rate of change of x wrt y.

When you wrote your formula, you treated both f^-1 and y as a function of the same variable x (or to be more accurate, functions of the same range), while it's not true.

To deliver this notion via the formula it's more accurate to write:$$\frac{d}{dy}f^{-1}(y)|_{y=y_{0}}=(\frac{d}{dx}f(x))^{-1}|_{x=f^{-1}(y_{0})}$$