107. Derivatives of Inverse Functions

Finding the derivative of the inverse function 𝑓−1(𝑥) at a given point directly from the function 𝑓(𝑥). Instead of explicitly computing the inverse function 𝑓−1(𝑥), we use the inverse function derivative formula:

𝑑𝑑𝑥[𝑓−1(𝑥)]=1𝑓′(𝑓−1(𝑥))

This approach allows us to determine the derivative of the inverse function without needing to express 𝑓−1(𝑥) explicitly. Instead, we find the value of 𝑥 that satisfies 𝑓(𝑥)=𝑎 (where a a is the given point), evaluate 𝑓′(𝑥), and apply the formula.

1. Definition of Inverse Function

𝑓(𝑓−1(𝑥))=𝑥𝑓−1(𝑓(𝑥))=𝑥

2. Differentiate Both Sides

Differenatiate both sides with respect to 𝑥

𝑑𝑑𝑥[𝑓(𝑓−1(𝑥))]=𝑑𝑑𝑥[𝑥]

The right-hand side simplifies to:

𝑑𝑑𝑥[𝑓(𝑓−1(𝑥))]=1

Using chain-rule:

𝑑𝑑𝑥[𝑓(𝑔(𝑥))]=𝑓′(𝑔(𝑥))⋅𝑔′(𝑥)

The left side expands as:

𝑑𝑑𝑥[𝑓(𝑓−1(𝑥))]=𝑓′(𝑓−1(𝑥))⋅𝑑𝑑𝑥[𝑓−1(𝑥)]

This we get:

𝑓′(𝑓−1(𝑥))⋅𝑑𝑑𝑥[𝑓−1(𝑥)]=1

3. Solve for 𝑑𝑑𝑥[𝑓−1(𝑥)]

Rearrange to isolate 𝑑𝑑𝑥[𝑓−1(𝑥)]:

𝑑𝑑𝑥[𝑓−1(𝑥)]=1𝑓′(𝑓−1(𝑥))
Example

Given the function:

𝑓(𝑥)=𝑥3

We want to find 𝑑𝑑𝑥𝑓−1(𝑥) at 𝑥=0.5 using the inverse function derivative formula:

𝑑𝑑𝑥𝑓−1(𝑥)=1𝑓′(𝑓−1(𝑥))

1. Compute 𝑓′(𝑥)

Differenatiate 𝑓(𝑥):

𝑓′(𝑥)=3𝑥2

2. Solve for 𝑥 such that 𝑓(𝑥)=0.5

We need to find 𝑥 such that:

𝑥3=0.5

Solving for 𝑥:

𝑥=0.53

3. Compute 𝑓′(0.53)

Evaluate the derivative at 𝑥=0.53:

𝑓′(0.53)=3(0.53)2

4. Use the formula

𝑑𝑑𝑥𝑓−1(0.5)=13(0.53)2

5. Interpretation

The expression:

𝑑𝑑𝑥𝑓−1(0.5)=13(0.53)2

represents the derivative of the inverse function 𝑓−1(𝑥) evaluated at 𝑥=0.5. This means it gives the slope of the tangent line to the inverse function at 𝑥=0.5.

inverse.py
from sympy import symbols, solve

# y = x**3 + x

x, y = symbols('x y')

f = x**3 + x - y

inverse = solve(f, x)

print(inverse)

107.1. Derivative Inverse Sin

𝑑𝑑𝑥[sin−1(𝑥)]=11−𝑥2𝑑𝑑𝑥[arcsin(𝑥)]=11−𝑥2

107.2. Derivative Inverse Cos

𝑑𝑑𝑥[cos−1(𝑥)]=−11−𝑥2𝑑𝑑𝑥[arccos(𝑥)]=−11−𝑥2

107.3. Derivative Inverse Tan

𝑑𝑑𝑥[tan−1(𝑥)]=−11+𝑥2𝑑𝑑𝑥[arctan(𝑥)]=−11+𝑥2