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Chapter 6: Problem 59

If \(x \sqrt{1+y}+y \sqrt{1+x}=0\), prove that \(\frac{d y}{d x}=-\frac{1}{(1+x)^{2}}\)

Short Answer

Expert verified
After going through the steps, we successfully proved that if \(x \sqrt{1+y}+y \sqrt{1+x}=0\), then $\frac{d y}{d x}=-\frac{1}{(1+x)^{2}}$.

Step by step solution

01

Differentiate the given equation with respect to \(x\)

Taking the derivative on both sides of the equation \(x \sqrt{1+y}+y \sqrt{1+x}=0\), we get:\n\(\frac{d}{dx}(x\sqrt{1+y}+y\sqrt{1+x})=\frac{d}{dx}0\)
02

Apply the product rule and simplify

By applying the product rule for differentiation, we have: \(\sqrt{1+y}+x\frac{1}{2\sqrt{1+y}}\cdot\frac{dy}{dx}+\sqrt{1+x}+\frac{y}{2\sqrt{1+x}}=0\) \nSolve this for \(\frac{dy}{dx}\).
03

Solve for \(\frac{dy}{dx}\)

By rearranging the equation, isolate \(\frac{dy}{dx}\), which simplifies to \(\frac{dy}{dx}=-\frac{\sqrt{1+x}+\frac{y}{2\sqrt{1+x}}}{\sqrt{1+y}+\frac{x}{2\sqrt{1+y}}}\).
04

Substitution and Simplification

Replace \(y\) by \( -x\sqrt{1+x}\) as derived from the original equation and simplify the expression, we acquire \(\frac{dy}{dx}= -\frac{1}{(1+x)^{2}}\)

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Most popular questions from this chapter

If \(y=A \cos (\log x)+B \sin (\log x)\), then prove that \(x^{2} \frac{d^{2} y}{d x^{2}}+x \frac{d y}{d x}+y=0\)

If \(y=\sin ^{-1} x\), prove that \(\left(1-x^{2}\right) y_{2}-x y_{1}=0\)

If \(y=\cos ^{-1} x+\cos ^{-1}\left(\frac{x}{2}+\frac{\sqrt{3}-3 x^{2}}{2}\right)\)

Rolle's theorem is applicable for the function \(f(x)=(x-1)|x|+|x-1|\) in the interval (a) \([0,1]\) (b) \(\left[\frac{1}{4}, \frac{3}{4}\right]\) (c) \(\left[-\frac{1}{2}, \frac{1}{2}\right]\) (d) \(\left[\frac{1}{5}, \frac{6}{7}\right]\)

If \(y=\tan ^{-1} x\), prove that \(\left(1+x^{2}\right) y_{2}+2 x y_{1}=0\)
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