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Chapter 6: Problem 66

If \(\tan ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=a\) prove that \(\frac{d y}{d x}=\frac{x(1-\tan a)}{y(1+\tan a)}\).

Short Answer

Expert verified
\frac{dy}{dx}= \frac{x(1-tan a)}{y(1+tan a)}

Step by step solution

01

Rewrite the given equation

Rewrite the given equation as \(\tan a=\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\).
02

Differentiate with respect to \(x\)

Differentiate the rewritten equation with respect to \(x\) using the chain rule. You鈥檒l get \(\sec^{2} a . da/dx= \frac{2x(x^{2}+y^{2})-(2x(x^{2}-y^{2}))}{(x^{2}+y^{2})^{2}}-\frac{2y(x^{2}+y^{2})-(2y(x^{2}-y^{2}))}{(x^{2}+y^{2})^{2}} \cdot dy/dx\).
03

Simplify the equation

Simplify the equation obtained from differentiation. You get \(da/dx=\frac{4xy(x^{2}+y^{2})}{(x^{2}+y^{2})^{2} \sec^{2} a - 2y . dy/dx}\). As \(\sec^{2} a= 1+\tan^{2} a\) replace sec^{2} a in above equation, we get \(da/dx=\frac{4xy(x^{2}+y^{2})}{((x^{2}+y^{2})^{2}(1+ \tan^{2} a) - 2y . dy/dx)}\).
04

Substitute for \(da/dx\) in original inverse tangent equation

Substitute for \(da/dx\) obtained in the original inverse tangent equation, the equation becomes \(\frac{dy}{dx}= \frac{1}{2y(1+\tan^{2}a)}-\frac{x(1-\tan a)}{y(1+\tan a)}\). As \(1+\tan^{2}a=\sec^{2}a\) replace in above equation which results in \(\frac{dy}{dx}= \frac{x(1-\tan a)}{y(1+\tan a)}\).
05

Conclusion

Hence the value of \(\frac{dy}{dx} =\frac{x(1-\tan a)}{y(1+\tan a)}\) has been derived successfully as required.

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