Finding an Inverse Derivative from an Implicit Input
To find the derivative of an inverse function at an output \(c\), we need the input that the original function sends to \(c\). Sometimes that input is not listed in a table or stated directly. It may need to be found by solving an equation, or by using a relation that defines the original function implicitly.
The key is to keep the roles of input and output clear. If \(f(a)=c\), then \(a=f^{-1}(c)\). The inverse derivative at \(c\) depends on the slope of \(f\) at \(a\), not on the slope at \(c\).
Two Steps: Recover the Input, Then Find the Slope
The calculation has two separate parts. First solve \(f(a)=c\) to identify the original input \(a\). Then find \(f'(a)\) and take its reciprocal. If \(f\) is given by an implicit relation rather than an explicit formula, implicit differentiation can provide the needed slope.
Solve \(f(a)=c\), or use the implicit relation to identify the point \((a,c)\) on the curve.
Calculate \(f'(a)\), using implicit differentiation if the function is defined by a relation.
Provided \(f'(a)\ne 0\), conclude \((f^{-1})'(c)=1/f'(a)\).
The output \(c\) becomes the input to the inverse function. The recovered value \(a\) is the input to the original function. This swap is why it is important to label the point before differentiating.
Why the Formula Works
Suppose \(g=f^{-1}\) near \(c\), with \(f(a)=c\). Under the conditions of the key formula, \(g\) is differentiable at \(c\); that is the content of the inverse function theorem, which is proved from the difference quotient \(\frac{g(y)-g(c)}{y-c}=\frac{g(y)-a}{f(g(y))-f(a)}\). Once differentiability of \(g\) is known, the derivative follows from the definition of an inverse: \(f(g(y))=y\) for \(y\) near \(c\). Differentiate this identity with respect to \(y\). The chain rule gives \(f'(g(y))g'(y)=1\). At \(y=c\), we have \(g(c)=a\), so \(f'(a)g'(c)=1\). If \(f'(a)\ne 0\), division gives the inverse derivative formula.
This argument explains both the reciprocal and the required condition: the original slope must not be zero, since otherwise the product \(f'(a)g'(c)=1\) could not hold. If a function is specified implicitly, we still need a point on its graph and its slope there. The inverse formula does not remove either requirement.
Worked Examples
Worked Example: Solve an Equation for the Original Input
Let \(f(x)=x^3+2x\). Find \((f^{-1})'(12)\).
First find the input \(a\) such that \(f(a)=12\):
Testing \(a=2\) gives \(2^3+2(2)=8+4=12\), so \(a=2\). Because \(f'(x)=3x^2+2>0\) for every real \(x\), \(f\) is strictly increasing, so \(a=2\) is the only solution and \(f\) has an inverse.
Since \(f'(2)=14\ne0\), the inverse derivative formula applies:
Check: \(f'(2)(f^{-1})'(12)=14\cdot\frac{1}{14}=1\).
Worked Example: Recover an Input and Use the Reciprocal
Let \(f(x)=x^3+x\). Find \((f^{-1})'(10)\).
Solve \(f(a)=10\). Since \(2^3+2=8+2=10\), the input is \(a=2\). Thus \(f(2)=10\), as required.
The original function is strictly increasing because \(f'(x)=3x^2+1>0\) for every real \(x\), so its inverse is defined. Since \(f'(2)=13\ne0\), the inverse derivative formula applies:
Check the inputs and slope: \(f(2)=10\), and \(f'(2)(f^{-1})'(10)=13\cdot\frac{1}{13}=1\), as required by the chain-rule identity.
Worked Example: Use an Implicit Curve to Find the Original Slope
Suppose \(y=f(x)\) is locally defined by \(x^2+xy+y^2=7\) near the point \((1,2)\). Find \((f^{-1})'(2)\).
The point tells us \(f(1)=2\), so the original input corresponding to the inverse input \(2\) is \(a=1\). Differentiate the relation with respect to \(x\), treating \(y\) as \(f(x)\):
Collect the terms containing \(y'\), then substitute \(x=1\) and \(y=2\):
The coefficient \(x+2y\) equals \(5\) at the point, so it is nonzero; the relation locally defines \(y\) as a function of \(x\). Also \(f'(1)=-\frac45\ne0\), so the inverse derivative exists locally. Therefore,
Check using the inverse relation. If \(x=g(y)=f^{-1}(y)\), then \(g(y)^2+g(y)\,y+y^2=7\). Differentiating with respect to \(y\) gives \(2g(y)g'(y)+g'(y)\,y+g(y)+2y=0\). At \(y=2\), \(g(2)=1\), so \(2g'(2)+2g'(2)+1+4=0\), that is \(4g'(2)+5=0\), and \(g'(2)=-\frac54\), confirming the answer.
Worked Example: Find the Input from an Implicit Equation
Near \((1,1)\), let \(y=f(x)\) be defined by \(y^3+xy=2\). Find \((f^{-1})'(1)\).
To find the original input that produces output \(1\), set \(y=1\) in the relation:
Thus \(f(1)=1\). Differentiate the relation with respect to \(x\):
At \(x=1,\ y=1\), this becomes \(4y'=-1\), so \(f'(1)=-\frac14\). The coefficient \(3y^2+x\) is \(4\ne0\), and the slope is nonzero; the inverse derivative formula applies:
For a check, write the local inverse as \(x=g(y)\) and differentiate \(y^3+g(y)y=2\). This gives \(3y^2+g'(y)y+g(y)=0\). At \(y=1,\ g(1)=1\), so \(3+g'(1)+1=0\), hence \(g'(1)=-4\).
Implicit Differentiation and the Inverse Slope
When a curve \(F(x,y)=0\) defines \(y=f(x)\) near \((a,c)\), implicit differentiation gives
If \(F_y(a,c)\ne0\), this yields \(f'(a)=-F_x(a,c)/F_y(a,c)\). If that slope is also nonzero, then
The final expression is valid under the stated conditions: the relation defines \(y\) locally as a function of \(x\), the original slope exists, and that slope is nonzero. In particular, \(F_x(a,c)\ne0\) is needed for the displayed reciprocal expression. This is a local calculation at a specified point; it does not provide a formula for the whole inverse function.
Common Mistake: Using the Output as the Original Input
A frequent error is to evaluate \(f'\) at \(c\), even though \(c\) is the input to the inverse, not the original function. For \(f(x)=x^3+x\), the equation \(f(a)=10\) gives \(a=2\). The incorrect calculation would be
But \(f(10)=1010\), not \(10\), so \(10\) is not the original input corresponding to the inverse input \(10\). The correction is to use \(a=2\):
A second useful check is that the two slopes at corresponding points must multiply to \(1\): \(f'(2)(f^{-1})'(10)=1\). If that product is not \(1\), recheck the recovered input and the derivative.
Why This Method Matters
An inverse function can be difficult to express explicitly, even when its derivative at one particular value is accessible. The inverse derivative formula lets us answer a local rate-of-change question using the original function’s slope. If the function is given implicitly, implicit differentiation supplies that slope without requiring us to solve the entire relation for \(y\).
The method is only as reliable as the input identification. Confirm that the point really satisfies the original relation, and check that the original derivative is defined and nonzero there. If several points on a relation could correspond to the same output, the relevant local inverse must be specified; determining which branch applies is a separate issue.
Check Your Understanding
For each problem, identify the original input before using the inverse derivative formula. State any condition needed for the formula to apply.
- If \(f(x)=x^3+2x\), solve \(f(a)=4\) and find \((f^{-1})'(4)\).
- If \(y=f(x)\) is defined near \((2,1)\) by \(x^2+y^2=5\), find \(f'(2)\) and \((f^{-1})'(1)\), if they exist.
- For \(y^3+xy=2\) near \((1,1)\), show how to identify the original input for output \(1\).
- Explain why \(f'(c)\) is generally not the slope to use when finding \((f^{-1})'(c)\).
- What happens to the inverse derivative formula if \(f'(a)=0\)?
