Find dy/dx: y = sin -1( 2 x √ (1 - x2), -1/√2 < x < 1/√2
Solution:
A derivative helps us to know the changing relationship between two variables. Consider the independent variable 'x' and the dependent variable 'y'.
The change in the value of the dependent variable with respect to the change in the value of the independent variable expression can be found using the derivative formula.
Given,
y = sin -1( 2 x √ (1 - x2)
Let us find the derivative on both sides with respect to x.
We will proceed by using the inverse trigonometric concept.
Let us assume x = sin θ
⇒ θ = sin-1 x. -----(1)
On substituting the value of x in the given function, we get
⇒ y = sin -1( 2 sin θ √ (1 - sin2 θ))
Using trigonometric identity, we get
⇒ y = sin -1( 2 sin θ cos θ)
⇒ y = sin -1( sin 2θ) = 2θ
= 2 sin-1 x [from equation 1]
On differentiating both sides wrt x, we get
dy/dx = 2 / √ (1 - x2)
Therefore,
the derivative of y = sin -1( 2 x √ (1 - x2) is 2 / √ (1 - x2)
NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.3 Question 14
Find dy/dx: y = sin -1( 2 x √ (1 - x2), -1/√2 < x < 1/√2
Summary:
The derivative of sin -1( 2 x √ (1 - x2), -1/√2 < x < 1/√2 with respect to x is dy/dx = 2 / √ (1 - x2) .A derivative helps us to know the changing relationship between two variables
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