Differentiate an inverse function using (f^-1)'(x) = 1/f'(f^-1(x)) (equivalently dy/dx = 1/(dx/dy)), including the second derivative

What this dot point is asking

NESA wants you to be able to differentiate the inverse of a function without first finding a formula for that inverse. The tool is one rule, written two equivalent ways:

This is the general result behind the inverse-trig derivatives you already use, and it powers the standard derivation of $\frac{d}{dx}(\ln x) = \frac{1}{x}$. The same dot point extends to the second derivative of an inverse, which is where most of the marks (and the mistakes) sit.

The skill examined is almost always evaluation at a point: you are given the value of $f$ and its derivatives at one input and asked for the gradient (or the second derivative) of the inverse at the matching output. You rarely need an explicit formula for $f^{-1}$, and reaching for one is usually the slow, error-prone route. A marker can tell instantly whether you understand that the inverse's gradient is read off the original function at the matching point, or are trying to invert the function by hand and differentiate that.

The answer

Why an inverse exists at all

A function has an inverse only if it is one-to-one: each output comes from exactly one input, so the rule can be run backwards unambiguously. The quickest sufficient test in this course is monotonicity: if $f'(x) > 0$ for all $x$ in the domain (strictly increasing) or $f'(x) < 0$ throughout (strictly decreasing), then $f$ is one-to-one and $f^{-1}$ exists. This matters here because the differentiation rule divides by $f'(f^{-1}(x))$, so it needs $f'$ to be non-zero at the relevant point. A horizontal tangent on $f$ ($f' = 0$) reflects into a vertical tangent on $f^{-1}$, where the inverse's gradient is undefined, exactly the division-by-zero the formula warns you about.

You usually do not have to invert the function. A one-line check that $f'$ keeps a constant sign (for example $f'(x) = 3x^2 + 1 \ge 1 > 0$) is enough to assert the inverse exists before you differentiate it.

The rule and where it comes from

There are two complementary derivations, and good "explain" or "show that" answers use whichever the question invites.

Chain rule on $f(f^{-1}(x)) = x$. By definition, applying $f$ to $f^{-1}(x)$ returns $x$. Differentiate both sides with respect to $x$, using the chain rule on the left:

valid wherever $f'(f^{-1}(x)) \neq 0$. This is the cleanest justification and the one to quote when a question says "show that".

The $dx/dy$ form (Leibniz). Write $y = f^{-1}(x)$, which is the same as $x = f(y)$. Differentiating $x = f(y)$ with respect to $y$ gives $\frac{dx}{dy} = f'(y)$, and treating the derivative as a ratio,

This is the form that appears on motion and related-rates questions, and it is identical to the first: $\frac{dx}{dy}$ evaluated at $y = g(x)$ is exactly $f'(g(x))$.

Reflection intuition. The graph of $f^{-1}$ is the graph of $f$ reflected in the line $y = x$. Reflection in $y = x$ swaps the horizontal and vertical directions, so it swaps the rise and the run of any tangent line. A tangent of gradient $\frac{\text{rise}}{\text{run}}$ on $f$ becomes a tangent of gradient $\frac{\text{run}}{\text{rise}}$ on $f^{-1}$, that is, the reciprocal. This is why the rule has a reciprocal in it, and it is the picture to draw if a question asks you to explain geometrically.

Picture: a curve, its inverse, and reciprocal tangents

The figure shows $y = x^3$ (its inverse is $y = \sqrt[3]{x}$, since cubing is one-to-one). At the point $P$ on the cubic the tangent has gradient $4$; the reflected point $P'$ on the inverse curve has a tangent of gradient $\frac{1}{4}$. The two tangents are themselves mirror images in $y = x$, which is the whole content of the rule.

Evaluating the gradient of an inverse at a point, stage by stage

The examined task is: given numbers about $f$ at one input, find $g'$ of $f^{-1}$ at the matching output. The danger is differentiating at the wrong point. The four panels below build the logic on $f(x) = x^3$ with the known point $(a, b)$.

Stage 1, start from the known point on $f$. You are told (or can compute) a point $(a, b)$ on $y = f(x)$, meaning $f(a) = b$. The matching point on the inverse is the reflection $(b, a)$, because $g(b) = a$. Keep $a$ and $b$ straight: $a$ is the input of $f$, $b$ is the output, and they swap roles for $g$.

Stage 2, read the gradient of $f$ at that point. Compute $f'(a)$. This single number is everything you need: it is the gradient of $f$ at $(a, b)$, the slope of the tangent shown.

Stage 3, reflect to the matching point on $g$. The inverse curve $y = g(x)$ is the reflection of $y = f(x)$ in $y = x$. The known point $(a, b)$ reflects to $(b, a)$, which lies on $g$. This is the point at which you want the inverse's gradient.

Stage 4, the inverse's gradient is the reciprocal. The tangent at $(b, a)$ on $g$ is the reflection of the tangent at $(a, b)$ on $f$, so its gradient is $\dfrac{1}{f'(a)}$. That is the answer: $g'(b) = \dfrac{1}{f'(a)}$, the reciprocal of the number from Stage 2.

So the whole evaluation is four moves: find the matching input $a$ (where $f(a)$ equals the value you want $g$ to act on), compute $f'(a)$, take the reciprocal. You never differentiate $g$ itself.

The second derivative of an inverse

This is the part Cambridge leaves as an exercise and the part HSC questions like to push into. Differentiate $g'(x) = \dfrac{1}{f'(g(x))} = \bigl[f'(g(x))\bigr]^{-1}$ using the chain rule. The outer power gives $-\bigl[f'(g(x))\bigr]^{-2}$, and the inner derivative of $f'(g(x))$ is $f''(g(x)) \cdot g'(x)$:

Now substitute $g'(x) = \dfrac{1}{f'(g(x))}$ for that last factor and combine the powers:

Two things are worth committing to memory because they are exactly the traps: the minus sign (so a positive $f''$ at the matching point gives a negative $g''$, and vice versa, which is the reflection flipping concavity), and the cube on $f'$ in the denominator (not a square, and not the reciprocal of $f''$). As with the first derivative, every piece is evaluated at the matching point $g(x)$, never at $x$.

Inverse-trig and log derivatives as special cases

The whole inverse-trig topic is this rule applied to $\sin$, $\cos$ and $\tan$. Take $f(x) = \sin x$ on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, so $g(x) = \arcsin x$ and $f'(x) = \cos x$. The rule gives

using $\cos(\arcsin x) = \sqrt{1 - x^2}$ on the principal range, which is precisely the implicit-differentiation derivation on the inverse-trig page, just packaged as the general rule. Likewise $f(x) = e^{x}$ has $g(x) = \ln x$ and $f'(x) = e^{x}$, so

Seeing these as the same rule is the point of the dot point: you are not learning a separate fact for every inverse, you are applying one reciprocal-of-the-derivative-at-the-matching-point idea.

How exam questions ask about differentiating an inverse

The wording tells you which form of the rule to deploy:

• "$f(a) = b$ and $f'(a) = m$. Find $g'(b)$" (table or values given). The core type. The matching point is $a$ (because $f(a) = b$ means $g(b) = a$), so $g'(b) = \frac{1}{f'(a)} = \frac{1}{m}$. No formula for $g$ needed.
• "$f(x) = \ldots$ Find $g'(c)$" (a concrete function). First solve $f(x) = c$ to find the matching input (often a neat integer by inspection), then take the reciprocal of $f'$ there. Confirm an inverse exists by checking $f'$ keeps one sign.
• "Find $g''(\ldots)$ given $f'$ and $f''$." Use $g''(x) = -\frac{f''(g(x))}{[f'(g(x))]^3}$, watching the sign and the cube. Compute $g(x)$ (the matching input) first, then substitute the supplied $f'$ and $f''$.
• "Find the equation of the tangent to $y = g(x)$ at $\ldots$". Find the point on $g$ (reflect the known point of $f$) and the gradient $\frac{1}{f'(a)}$, then use point-gradient form.
• "Explain geometrically why the gradients are reciprocals" / "using reflection in $y = x$". A graph answer: reflection swaps rise and run, so the gradient inverts. Draw the curve, its mirror image, and the two tangents.
• "Show that $\frac{dy}{dx} = \frac{1}{dx/dy}$" or "by writing $x = f(y)$". A derivation: differentiate $f(f^{-1}(x)) = x$ by the chain rule, or differentiate $x = f(y)$ with respect to $y$ and invert. State where it fails ($f' = 0$, i.e. vertical tangent on $g$).