Relation between Derivatives of Inverse Functions
As we already know that what is inverse function, so now we are going to find the derivative of inverse function. So, if a function f(x) is a continuous one-to-one function or bijective function defined on an interval lets say I, then its inverse is also continuous and if the function f(x) is a differentiable function, then its inverse is also a differentiable function.
g'(x) = [Tex]\mathbf{\frac{1}{f'(g(x))}} [/Tex]
Here, f and g are inverse function. It is known as inverse function theorem.
Proof:
Let us considered f and g be the inverse functions and x is present in the g domain, then
f(g(x)) = x
Differentiate both side with respect to x
[Tex]\frac{d f(g(x))}{x} = \frac{d(x)}{dx}[/Tex]
Now solve the LHS using chain rule we get
f'(g(x))g'(x) = 1
Now solve for g'(x):
g'(x) = [Tex]{\frac{1}{f'(g(x))}} [/Tex]
Hence, the derivative on inverse function is solved.
Example 1: f(x) = ex, check whether condition holds true.
Solution:
As, f(x) = ex
y = ex
x = ln y
g(x) = f-1(x) = ln x
Now,
f'(x) = [Tex]\frac{d}{dx} [/Tex] (ex) = ex
g'(x) = [Tex]\frac{d}{dx} [/Tex] (ln x) = [Tex]\frac{1}{x}[/Tex]
g'(f(x)) = [Tex]\frac{1}{e^x}[/Tex]
[Tex]\frac{1}{g'(f(x))} = \frac{1}{\frac{1}{e^x}} [/Tex] = ex
Hence,
f'(x) = [Tex]\frac{1}{g'(f(x))} [/Tex], holds true.
Example 2: Let f(x) = [Tex]\frac{1}{2} [/Tex]x3 + 3x – 4, and let g be the inverse function of f where f(-2) = -14. Find g'(-14)
Solution:
f'(x) = [Tex]\frac{1}{2} [/Tex] (3x2) + 3
According to eq (1).
[Tex]g'(x) = \mathbf{\frac{1}{f'(g(x))}}[/Tex]
As, f(x) = g-1(x) and f(-2) = -14
then g(-14) = -2
x = -14
g'(-14) = [Tex]\frac{1}{f'(g(-14))}[/Tex]
g'(-14) = [Tex]\frac{1}{f'(-2)}[/Tex]
f'(-2) = [Tex]\frac{1}{2} [/Tex] (3(-2)2) + 3
f'(-2) = [Tex]\frac{12}{2} [/Tex]+ 3
f'(-2) = 9
then, g'(-14) = [Tex]\frac{1}{9}[/Tex]
Derivatives of Inverse Functions
In mathematics, a function(e.g. f), is said to be an inverse of another(e.g. g), if given the output of g returns the input value given to f. Additionally, this must hold true for every element in the domain co-domain(range) of g. E.g. assuming x and y are constants if g(x) = y and f(y) = x then the function f is said to be an inverse of the function g. Or in other words, if a function f : A ⇢ B is one – one and onto function or bijective function, then a function defined by g : B ⇢ A is known as inverse of function f. The inverse function is also known as the anti function. The inverse of function is denoted by f-1.
f(g(x)) = g(f(x)) = x
Here, f and g are inverse functions.
Table of Content
- Overview of Derivatives of Inverse Functions
- Procedure of finding inverse of f
- Derivatives of Inverse Functions
- How to find derivatives of inverse functions from the table?
- Derivatives of Inverse Trigonometric Functions
- How to find the derivatives of inverse trigonometric functions?
- 1. Derivative of f given by f(x) = sin–1 x.
- 2. Derivative of f given by f(x) = cos–1 x.
- 3. Derivative of f given by f(x) = tan–1 x.
- 4. Derivative of f given by f(x) = cot–1 x.
- 5. Derivative of f given by f(x) = sec–1 x.
- 6. Derivative of f given by f(x) = cosec–1 x.
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