Reading an Inverse Derivative from a Graph
A graph can show more than the values of a function. At a point, it can also show the function’s instantaneous rate of change: the slope of the tangent line. For an inverse function, the key is to match the point correctly and then use the relationship between the two tangent slopes.
The graph of an inverse function is obtained by interchanging the input and output coordinates. Consequently, a point \((a,b)\) on the graph of \(f\) corresponds to the point \((b,a)\) on the graph of \(f^{-1}\). At corresponding points, the tangent slopes are reciprocals, provided the original tangent slope is nonzero and the inverse derivative exists.
Match the Input Before Calculating
The input to \((f^{-1})'\) is an output of \(f\). If the graph of \(f\) shows the point \((a,b)\), then \(f(a)=b\). The inverse takes \(b\) back to \(a\), so the inverse graph contains \((b,a)\), and its derivative at that point is \((f^{-1})'(b)\).
This coordinate swap is easy to overlook. A request for \((f^{-1})'(b)\) means that we need the point on the original graph whose output coordinate is \(b\), not necessarily the point whose input coordinate is \(b\). Once the corresponding original point \((a,b)\) is identified, read the tangent slope there.
Reading a Tangent Slope Accurately
A graph’s tangent slope is not generally the slope of a nearby secant line. It is the slope of the line that best represents the graph’s direction at the point. If the tangent is drawn, choose two convenient points on that tangent—not necessarily points on the curve—and calculate rise divided by run.
Use the numerical axis scales, not simply the number of grid squares. For example, if one vertical square represents three units and one horizontal square represents two units, a rise of three squares and a run of two squares correspond to a vertical change of nine and a horizontal change of four.
After finding the original tangent slope \(m\), take its reciprocal, \(1/m\), rather than changing its sign. A negative original slope produces a negative reciprocal; the sign does not disappear.
The Inverse Derivative Rule and Its Conditions
For this graph method, the relevant result is the inverse derivative formula. Its conditions matter: there must be an inverse on an interval around the original input, and the inverse must approach the corresponding original input as its input approaches the output value. One standard set of sufficient conditions is continuity and one-to-one behavior of \(f\) on an interval containing \(a\), together with differentiability at \(a\) and a nonzero derivative there.
The theorem explains why a nonzero tangent slope can be reversed by taking its reciprocal. Continuity and one-to-one behavior ensure that \(f^{-1}\) is locally well-defined and that its input values close to \(b\) correspond to original inputs close to \(a\). The nonzero derivative makes the quotient in the proof meaningful.
Why the Reciprocal Appears
The coordinate swap gives the point correspondence, but the derivative formula can also be justified directly with a difference quotient. Write \(g=f^{-1}\), so \(g(b)=a\). For \(y\ne b\) sufficiently close to \(b\), set \(x=g(y)\). Then \(f(x)=y\). Since \(g\) is continuous at \(b\), \(y\to b\) implies \(x=g(y)\to a\).
As \(y\to b\), the corresponding \(x\to a\). Differentiability of \(f\) at \(a\) means that the denominator in the rightmost expression approaches \(f'(a)\). Because \(f'(a)\ne 0\), taking reciprocals is valid, and the limit is \(1/f'(a)\). This is exactly the derivative of \(g\) at \(b\).
Geometrically, reflection across \(y=x\) swaps horizontal and vertical changes. If a tangent to the original graph has nonzero slope \(\Delta y/\Delta x\), the reflected tangent has slope \(\Delta x/\Delta y\). This geometric explanation agrees with the difference-quotient proof.
A Reliable Graph Procedure
For a requested derivative \((f^{-1})'(b)\), find the point \((a,b)\) on the graph of \(f\). Check that its output coordinate is \(b\).
The corresponding point on the inverse graph is \((b,a)\). This identifies where the inverse derivative is being evaluated.
Use two points on the tangent and account for the graph’s axis scales to calculate \(f'(a)\).
Confirm the slope is nonzero and the inverse is valid locally. Then calculate \((f^{-1})'(b)=1/f'(a)\).
Worked Examples
Worked Example: A Positive Tangent Slope
A graph of \(f\) shows the point \((2,5)\). Its tangent line rises 3 units for every 2 units it moves to the right. Find \((f^{-1})'(5)\), assuming \(f\) has a differentiable local inverse there.
The point \((2,5)\) says \(f(2)=5\). Therefore, the inverse graph contains the reflected point \((5,2)\), and the requested derivative is the inverse slope at input \(5\).
The tangent slope on the original graph is \[ f'(2)=\frac{3}{2}. \] It is nonzero, so the reciprocal rule applies under the stated local-inverse assumption: \[ (f^{-1})'(5)=\frac{1}{f'(2)} =\frac{1}{3/2} =\frac{2}{3}. \]
Check: the product of the corresponding tangent slopes is \((3/2)(2/3)=1\), as required. The answer is at the reflected point \((5,2)\), not at the original point \((2,5)\).
Worked Example: A Negative Tangent Slope
At the point \((-1,4)\), a graph of \(f\) has a tangent that falls 2 units while moving 5 units to the right. Find \((f^{-1})'(4)\), assuming the inverse is differentiable at \(4\).
The original point gives \(f(-1)=4\), so the inverse graph contains \((4,-1)\). The slope on the original graph is negative: \[ f'(-1)=\frac{-2}{5}=-\frac{2}{5}. \]
Take the reciprocal, retaining the negative sign: \[ (f^{-1})'(4)=\frac{1}{f'(-1)} =\frac{1}{-2/5} =-\frac{5}{2}. \]
Check: \((-2/5)(-5/2)=1\). Both corresponding tangents slope downward, and their slopes have reciprocal magnitudes.
Worked Example: Using the Graph’s Axis Scales
A graph has horizontal tick marks 2 units apart and vertical tick marks 3 units apart. At the point \((6,-2)\), the tangent moves right 2 grid squares and up 3 grid squares. Find \((f^{-1})'(-2)\), assuming a local differentiable inverse exists.
First convert grid squares to coordinate changes. The horizontal change is \(2\cdot 2=4\) units, while the vertical change is \(3\cdot 3=9\) units. Thus \[ f'(6)=\frac{9}{4}. \]
Since \(f(6)=-2\), the inverse graph contains the reflected point \((-2,6)\). The requested inverse derivative is \[ (f^{-1})'(-2)=\frac{1}{f'(6)} =\frac{1}{9/4} =\frac{4}{9}. \]
Check: \((9/4)(4/9)=1\). The scale conversion is essential; using the grid-square ratio \(3/2\) would give the wrong original slope and therefore the wrong inverse derivative.
Common Mistakes and What the Graph Can Establish
A frequent error is to use the original slope instead of its reciprocal. For example, if a graph gives \(f'(2)=3/2\) and \(f(2)=5\), writing \((f^{-1})'(5)=3/2\) is incorrect. The inverse interchanges the horizontal and vertical changes, so the correct result is \(2/3\). The check \((3/2)(2/3)=1\) helps catch this error.
Another error is to evaluate at the wrong coordinate. From \(f(2)=5\), the inverse derivative is evaluated at \(5\), not \(2\): the original point \((2,5)\) becomes \((5,2)\). Always identify the requested inverse input before reading the original point.
A graph’s scales also matter. Counting squares without reading axis increments can change the slope. Moreover, a horizontal tangent has slope \(0\), whose reciprocal is undefined. The inverse derivative formula does not apply when \(f'(a)=0\); the graph alone does not justify reporting a finite inverse derivative by taking a reciprocal.
Check Your Understanding
For each question, identify the reflected point as well as the derivative value. Assume the inverse exists and is differentiable wherever the question asks for an inverse derivative.
- A graph shows \(f(3)=7\) and \(f'(3)=4\). What is \((f^{-1})'(7)\), and where is the corresponding inverse point?
- At \((-2,6)\), the tangent slope to \(f\) is \(-3\). Find \((f^{-1})'(6)\).
- The original tangent rises 4 units while moving right 6 units. What is the corresponding inverse tangent slope?
- Why must you use axis scales rather than only count grid squares when reading a tangent slope?
- If \(f'(a)=0\), why can the reciprocal-slope rule not be used to claim a finite inverse derivative at \(f(a)\)?
