Find Derivative Of Inverse Function At A Point Calculator

This calculator helps you find the derivative of the inverse function f⁻¹(x) at a specific point ‘a’, given that f(b)=a and you know the value of f'(b).

What is the Derivative of an Inverse Function at a Point?

The derivative of an inverse function at a point ‘a’, denoted as (f⁻¹)'(a) or g'(a) where g(x) = f⁻¹(x), tells us the rate of change of the inverse function f⁻¹(x) with respect to x at the point x=a. If we have a differentiable and invertible function f(x), and its inverse f⁻¹(x), the derivative of the inverse at ‘a’ is related to the derivative of the original function at a corresponding point ‘b’ where f(b) = a.

Essentially, if the graph of y=f(x) has a tangent line at (b, a) with slope f'(b), then the graph of y=f⁻¹(x) (which is a reflection of y=f(x) across y=x) will have a tangent line at (a, b) with a slope that is the reciprocal of f'(b), provided f'(b) is not zero. The derivative of inverse function at a point calculator helps you find this reciprocal slope.

This concept is useful in calculus when we have an inverse function but its explicit formula is difficult or impossible to find, yet we know the derivative of the original function. It’s used by students, engineers, and scientists working with function transformations and rates of change. A common misconception is that the derivative of the inverse is the inverse of the derivative, which is incorrect; it’s the reciprocal of the derivative of the original function at the corresponding point.

Derivative of Inverse Function at a Point Formula and Mathematical Explanation

Let y = f(x) be a differentiable and invertible function, and let x = g(y) = f⁻¹(y) be its inverse. We want to find the derivative of g(y) with respect to y, which is g'(y) or (f⁻¹)'(y).

We know that if g(y) = f⁻¹(y), then f(g(y)) = y.

Differentiating both sides with respect to y, using the chain rule on the left side, we get:

f'(g(y)) * g'(y) = 1

Solving for g'(y), we find:

g'(y) = 1 / f'(g(y))

If we want to find the derivative of the inverse function at a specific point, say y = a, we have:

g'(a) = (f⁻¹)'(a) = 1 / f'(g(a))

Now, let’s say g(a) = b. This means f(b) = a. So, we can rewrite the formula as:

(f⁻¹)'(a) = 1 / f'(b), where f(b) = a.

This is the formula our derivative of inverse function at a point calculator uses. We need the point ‘a’, the corresponding value ‘b’ such that f(b)=a, and the value of the derivative f'(b).

Variables Table

Variable	Meaning	Unit	Typical Range
a	The point at which we find the derivative of the inverse f⁻¹(x)	(Unit of f(x))	Real numbers
b	The point such that f(b) = a	(Unit of x)	Real numbers
f'(b)	The derivative of f(x) evaluated at x=b	(Unit of f(x)/Unit of x)	Real numbers, not zero
(f⁻¹)'(a)	The derivative of f⁻¹(x) evaluated at x=a	(Unit of x/Unit of f(x))	Real numbers

Table of variables for the derivative of the inverse function formula.

Practical Examples (Real-World Use Cases)

Let’s look at some examples of using the derivative of inverse function at a point calculator.

Example 1: f(x) = x³ + x

Suppose f(x) = x³ + x. We want to find the derivative of its inverse f⁻¹(x) at a=2.

First, we need to find ‘b’ such that f(b) = 2. So, b³ + b = 2. By inspection, b=1 works (1³ + 1 = 2).

Next, we find the derivative of f(x): f'(x) = 3x² + 1.

Now, we evaluate f'(b) = f'(1) = 3(1)² + 1 = 4.

Using the formula (f⁻¹)'(a) = 1 / f'(b), we get:

(f⁻¹)'(2) = 1 / f'(1) = 1 / 4 = 0.25

So, the derivative of the inverse of f(x) = x³ + x at a=2 is 0.25.

Example 2: f(x) = e^x

Let f(x) = e^x. Its inverse is f⁻¹(x) = ln(x). We want to find the derivative of f⁻¹(x) at a=e².

We need ‘b’ such that f(b) = e^b = e². So, b=2.

The derivative is f'(x) = e^x. At b=2, f'(b) = f'(2) = e².

Using the formula: (f⁻¹)'(a) = 1 / f'(b)

(f⁻¹)'(e²) = 1 / f'(2) = 1 / e².

We know f⁻¹(x) = ln(x), so (f⁻¹)'(x) = 1/x. At x=e², (f⁻¹)'(e²) = 1/e², which matches.

The derivative of inverse function at a point is a powerful tool even when we don’t explicitly know the inverse function, as seen in Example 1 before recognizing b.

How to Use This Derivative of Inverse Function at a Point Calculator

Using the calculator is straightforward:

1. Enter Point ‘a’: In the first input field, enter the value ‘a’ at which you want to calculate the derivative of the inverse function, (f⁻¹)'(a).
2. Enter Value ‘b’: In the second field, enter the value ‘b’ for which f(b) = a. You might need to solve f(b)=a for b before using the calculator if you only have f(x) and ‘a’.
3. Enter f'(b): In the third field, enter the value of the derivative of the original function f(x) evaluated at x=b, i.e., f'(b). You’ll need to find f'(x) and then substitute x=b.
4. View Results: The calculator automatically updates the “Primary Result” showing the value of (f⁻¹)'(a) and also displays the intermediate values you entered.
5. Reset: Use the “Reset” button to clear the inputs to their default values.
6. Copy: Use the “Copy Results” button to copy the primary result and intermediate values to your clipboard.

The calculator assumes you have already found ‘b’ and calculated f'(b). The derivative of inverse function at a point is then simply 1/f'(b).

Key Factors That Affect Derivative of Inverse Function at a Point Results

The value of (f⁻¹)'(a) is directly influenced by the value of f'(b):

1. Magnitude of f'(b): The larger the absolute value of f'(b), the smaller the absolute value of (f⁻¹)'(a) because it’s the reciprocal. A steep slope on f at (b,a) means a shallow slope on f⁻¹ at (a,b).
2. Sign of f'(b): If f'(b) is positive, (f⁻¹)'(a) is also positive. If f'(b) is negative, (f⁻¹)'(a) is also negative. The original function and its inverse are either both increasing or both decreasing at corresponding points.
3. f'(b) close to zero: If f'(b) is very close to zero (but not zero), (f⁻¹)'(a) will be very large in magnitude. This corresponds to a nearly horizontal tangent on f and a nearly vertical tangent on f⁻¹.
4. f'(b) being zero: The formula is undefined if f'(b)=0. This corresponds to a horizontal tangent on f at x=b, meaning the inverse function would have a vertical tangent at x=a and is not differentiable there.
5. The function f(x): The nature of the function f(x) determines f'(x) and thus f'(b), which in turn determines (f⁻¹)'(a).
6. The points a and b: The specific values of a and b (where f(b)=a) determine at which point the derivative f'(b) is evaluated, affecting the final result for the derivative of inverse function at a point ‘a’.

Frequently Asked Questions (FAQ)

What is the inverse function theorem derivative?

The formula (f⁻¹)'(a) = 1 / f'(b) where f(b)=a is a direct result of the inverse function theorem, which relates the derivative of a function to the derivative of its inverse.

Why do we need f(b)=a?

The point (b, a) on the graph of f corresponds to the point (a, b) on the graph of f⁻¹. The derivative of f⁻¹ at ‘a’ is related to the derivative of f at ‘b’.

What if f'(b) = 0?

If f'(b) = 0, the formula 1/f'(b) is undefined. This means the tangent to f at x=b is horizontal, and the tangent to f⁻¹ at x=a would be vertical, so f⁻¹ is not differentiable at a.

Can I use this calculator if I only know f(x) and ‘a’?

Yes, but you first need to solve f(b) = a for ‘b’ and then find f'(x) and evaluate f'(b). The calculator itself needs ‘a’, ‘b’, and ‘f'(b)’.

How to find derivative of inverse function if f(x) is given?

1. Set f(b) = a to find b in terms of a (if possible). 2. Find f'(x). 3. Evaluate f'(b). 4. Calculate 1/f'(b) to get the derivative of inverse function at a point ‘a’.

Is the derivative of f inverse always 1 divided by f prime?

It’s 1 divided by f prime evaluated at the corresponding point, f⁻¹(a) or b, i.e., (f⁻¹)'(a) = 1 / f'(f⁻¹(a)).

Does this work for trigonometric inverse functions?

Yes, for example, if f(x)=sin(x), then f⁻¹(x)=arcsin(x). If a=1/2, b=π/6 (sin(π/6)=1/2), f'(x)=cos(x), f'(π/6)=√3/2. So (arcsin)'(1/2) = 1/(√3/2) = 2/√3, which is correct.

What if f is not one-to-one?

If f is not one-to-one, it doesn’t have a unique inverse function over its entire domain. You might need to restrict the domain of f to make it one-to-one before finding the derivative of the inverse for that restricted domain.