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

1. Feb 4, 2010

### 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 $$( f^{-1} )'(x) = \frac{1}{ \frac{dy}{dx} } = \frac{1}{3x^2+4x}?$$

2. Feb 4, 2010

### JSuarez

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

3. Feb 5, 2010

### Juwane

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.

4. Feb 5, 2010

### 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:

$$\frac{d}{dy}f^{-1}(y)|_{y=y_{0}}=(\frac{d}{dx}f(x))^{-1}|_{x=f^{-1}(y_{0})}$$