\(x\sqrt{1+y}+y\sqrt{1+x}=0\), for -1<x<1,prove that \(\frac{dy}{dx}\)=\(-\)\(\frac{1}{(1+x)^2}\)
\(x\sqrt{1+y}+y\sqrt{1+x}=0\), for -1<x<1,prove that \(\frac{dy}{dx}\)=\(-\)\(\frac{1}{(1+x)^2}\)
Solution and Explanation
It is given that
\(x\sqrt{1+y}+y\sqrt{1+x}=0\)
⇒ \(x\sqrt{1+y}=-y\sqrt{1+x}\)
squaring both sides, we obtain x2(1+y)=y2(1+x)
⇒ x2+x2y=y2+xy2
⇒ x2-y2=xy2-x2y
⇒ x2-y2=xy(y-x)
⇒ (x+y)(x-y)=xy(y-x)
∴x+y=-xy
⇒ (1+x)y=-x
⇒ y=-x/(1+x)
Differentiating both sides with respect to x, we obtain
\(\frac{dy}{dx}=(1+x)\frac{d}{dx}(x)-x\frac{d}{dx}\frac{(1+x)}{(1+x)^2}=-(1+x)-\frac{x}{(1+x)^2}\)
=\(-\frac{1}{(1+x)^2}\)
Hence, it proved.
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Notes on Continuity and differentiability
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.