Proof of Derivative of Inverse Trig Functions
We can differentiate the inverse trigonometric functions using the first principle and also by using implicit differentiation formula which also involves the use of chain rule. To find the derivative of inverse trigonometric functions using first principle is a lengthy process. In this article we learn how to differentiate inverse trigonometric functions using implicit differentiation. We can find the derivative (dy/dx) of inverse trig functions using following steps
Step 1: Assume the trigonometric functions in the form sin y = x
Step 2: Find the derivative of above function using implicit differentiation
Step 3: Calculate dy/dx
Step 4: Replace the value of trigonometric function present in the step 3 using trigonometric identities.
Derivative of sin inverse x
Let us assume sin y = x
Differentiating both side with respect to x
⇒ cos y. dy/dx = 1
⇒ dy/dx = 1/cos y →(i)
Since we know that Sin2y + Cos2y = 1
⇒ Cos2y = 1 – sin2y
⇒ cosy = √(1 – sin2y) = √(1 – x2) as we have sin y = x
Putting this value of cos y in equation (i)
dy/dx = 1/√(1 – x2) where y = sin-1x
Derivative of cos inverse X
Let us assume cos y = x
Differentiating both side with respect to x
⇒ -sin y. dy/dx = 1
⇒ dy/dx = -1/sin y →(i)
Since we know that Sin2y + Cos2y = 1
⇒ sin2y = 1 – cos2y
⇒ sin y = √(1 – cos2y) = √(1 – x2) as we have cos y = x
Putting this value of sin y in equation (i)
dy/dx = -1/√(1 – x2) where y = cos-1x
Derivative of tan inverse X
Let us assume tan y = x
Differentiating both side with respect to x
⇒ sec2y. dy/dx = 1
⇒ dy/dx = 1/sec2y →(i)
Since we know that sec2y – tan2y = 1
⇒ sec2y = 1 + tan2y
⇒ sec2y = (1 + tan2y) = (1 + x2) as we have tan y = x
Putting this value of sec2y in equation (i)
dy/dx = 1/(1 + x2) where y = tan-1x
Derivative of cot inverse X
Let us assume cot y = x
Differentiating both side with respect to x
⇒ -cosec2 y. dy/dx = 1
⇒ dy/dx = -1/cosec2y →(i)
Since we know that csec2y – cot2y = 1
⇒ cosec2y = 1 + cot2y
⇒ cosec2y = (1 + cot2y) = (1 + x2) as we have cot y = x
Putting this value of cosec2y in equation (i)
dy/dx = -1/(1 + x2) where y = cot-1x
Derivative of sec inverse X
Let us assume sec y = x
Differentiating both side with respect to x
⇒ sec y.tan y.dy/dx = 1
⇒ dy/dx = 1/sec y.tan y →(i)
Since we know that sec2y – tan2y = 1
⇒ tan2y = sec2y – 1
⇒ tan y = √(sec2y – 1) = √(x2 – 1)as we have sec y = x
Putting this value of tan y in equation (i)
dy/dx = 1/{|x|√(x2 – 1)} where sec y = x and y = sec-1x
Derivative of cosec inverse X
Let us assume cosec y = x
Differentiating both side with respect to x
⇒ -cosec y.cot y.dy/dx = 1
⇒ dy/dx = -1/cosec y.cot y →(i)
Since we know that cosec2y – cot2y = 1
⇒ cot2y = cosec2y – 1
⇒ cot y = √(cosec2y – 1) = √(x2 – 1)as we have cosec y = x
Putting this value of tan y in equation (i)
dy/dx = -1/{|x|√(x2 – 1)} where cosec y = x and y = cosec-1x
Derivative of Inverse Trig Functions
Derivative of Inverse Trig Function refers to the rate of change in Inverse Trigonometric Functions. We know that the derivative of a function is the rate of change in a function with respect to the independent variable. Before learning this, one should know the formulas of differentiation of Trigonometric Functions. To find the derivative of the Inverse Trigonometric Function, we will first equate the trigonometric function with another variable to find its inverse and then differentiate it using the implicit differentiation formula.
In this article, we will learn the Derivative of Inverse Trig Functions, Formulas of Differentiation of Inverse Trig Functions, and Solve some Examples based on it. But before heading forward, let’s brush up on the concept of inverse trigonometric functions and implicit differentiation.
Table of Content
- Inverse Trigonometric Functions
- What is Implicit Differentiation?
- What is Derivative of Inverse Trigonometric Functions?
- Proof of Derivative of Inverse Trig Functions
- Inverse Trig Derivative Formula
- Inverse Trig Derivative Examples