# Why is the inverse of the derivative of f not the actual derivative of the inverse of f?

So, I've explored this a little, but it is still confusing. When you calculate the inverse of a function, f, that is one-to-one, the points switch: a point (2,8) on f would be (8,2) on the inverse. So, one would assume that the derivatives of the functions would also constitute the reversal of points. However, that is not the case. For example, you have:

$f (x) = 5x^2 \phantom{=}\text{ for$x\geq0$}$

$f '(x) = 10x$

and

$(f^{-1}) (x) = \sqrt{\frac{x}{5}}$

Here is my question: Why is finding the inverse of the derivative of $f$, $f '(x)$, and taking its inverse not the real derivative of the inverse? I would think $(f^{-1}) '(x) = \frac{x}{10}$, but that is not the case. The real inverse would be taking the derivative of $(f^{-1}) (x)$ and finding $(f^{-1}) '(x) = (\frac{1}{10\sqrt{x/5}})$. In my mind, both of these seem like they could be the derivatives of the inverse, yet only the latter is true. Why is this?

Also, maybe I missed out in class, but is there some sort of quick relationship between (besides the formula) $f '(x)$ and $(f^{-1}) '(x)$ similar to how points switch between $f (x)$ and $(f^{-1}) (x)$.

Thanks.

• Hint: Try using the chain rule on the equation $f \circ f^{-1}(x) = x$. – Jair Taylor Jan 23 '18 at 18:21
• Welcome to MathSE. To typeset math, you can put dollar signs around an expression: $f(x)=5x^2$ gives $f(x)=5x^2$. – Théophile Jan 23 '18 at 18:44

Intuitive thoughts to reflect on: draw the graph of $f$ and mark a point on it (say $(a,f(a))$). Draw the tangent at that point. It will have slope $f'(a)$.

Now flip the entire plane around the line $y=x$. The graph of $f$ has now become the graph of $f^{-1}$, the marked point has become $$(f(a),a)= (f(a),f^{-1}(f(a)))=(b,f^{-1}(b))$$ where $b=f(a)$. The tangent line is still the tangent line, but its slope is inverted ($\Delta y$ and $\Delta x$ have swapped roles for the line, so their ratio is inverted).

Putting this together, we get $f'(a)=\frac{1}{f^{-1}(b)}$. The fact that $a$ and $b$ both appear here is what makes the expressions for $f'(x)$ and $(f^{-1})'(x)$ look less related than they are. Geometrically, the derivative of $f$ and of $f'$ at the same point in the plane (allowing for flipping the plane, of course) are very related. Algebraically, the derivative of $f$ and of $f'$ at the same input value are less so.

Maybe this isn't a great reason, but why should you expect the inverse of the derivative to be the derivative of the inverse? Take some simple examples to see why it shouldn't be a universal truth.

The function $f(x) = x$ has derivative $f'(x)=1$, which isn't even invertible. And there are hardly nicer functions than $f(x)=x$.

For your last question, note that $f(f^{-1}(x))=x$ by definition (when $f$ is invertible). Then, assuming enough differentiability, the chain rule gives $$\frac{d}{dx} f(f^{-1}(x)) = f'(f^{-1}(x))\cdot \frac{d}{dx}f^{-1}(x) = 1.$$ Hence $$\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}.$$ This is what you need.

Other answerers have given nice geometric reasons why this should be the case.