Find \(\frac{dy}{dx}\), if y=sin-1x+sin-1\(\sqrt{1-x^2}\), -1≤x≤1
Find \(\frac{dy}{dx}\), if y=sin-1x+sin-1\(\sqrt{1-x^2}\), -1≤x≤1
Solution and Explanation
It is given that ,y=sin-1x+sin-1\(\sqrt{1-x^2}\)
∴\(\frac{dy}{dx}\)=\(\frac{d}{dx}\)(sin-1x+sin-1\(\sqrt{1-x^2}\))
\(\Rightarrow\) \(\frac{dy}{dx}\)=\(\frac{d}{dx}\)(sin-1x)+\(\frac{d}{dx}\)(sin-1\(\sqrt{1-x^2}\))
\(\Rightarrow\) \(\frac{dy}{dx}\)=\(\frac{1}{\sqrt{1-x^2}}\)+\(\frac{1}{1-(\sqrt{1-x^2})^2}\).\(\frac{d}{dx}\)(\(\sqrt{1-x^2}\))
\(\Rightarrow\) \(\frac{dy}{dx}\)=\(\frac{1}{\sqrt{1-x^2}}\)+\(\frac{1}{x}\).\(\frac{1}{2}\)\(\sqrt{1-x^2}\).\(\frac{d}{dx}\)(1-x2)
\(\Rightarrow\) \(\frac{dy}{dx}\)=\(\frac{1}{\sqrt{1-x^2}}\)+\(\frac{1}{2x\sqrt{1-x^2}(-2x)}\)
\(\Rightarrow\) \(\frac{dy}{dx}\)=\(\frac{1}{\sqrt{1-x^2}}\)-\(\frac{1}{\sqrt{1-x^2}}\)
∴\(\frac{dy}{dx}\)=0
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.