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Chapter 6: Problem 124

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

Short Answer

Expert verified
After caclulating the first derivative \(y_{1} = \frac{1}{1+x^2}\) and the second derivative \(y_{2} = \frac{-2x}{(1+x^2)^2}\), substituting these into the given equation \(\left(1+x^{2}\right) y_{2}+2 x y_{1}=0\) results in zero, thereby proving the equation.

Step by step solution

01

Calculate First Derivative

Solve the first derivative of \(y = \tan^{-1} x\). By recognizing the function as an inverse trigonometric function, we know the derivative of \( \tan^{-1} x\) is \(\frac{1}{1+x^2}\). So \(y_{1} = \frac{1}{1+x^2}\).
02

Calculate Second Derivative

To obtain the second derivative \(y_{2}\), differentiate \(y_{1} = \frac{1}{1+x^2}\) implicitly. The derivative of \(\frac{1}{1+x^2}\) is \(\frac{-2x}{(1+x^2)^2}\). So \(y_{2} = \frac{-2x}{(1+x^2)^2}\).
03

Substitution into Equation

Substitute \(y_{1}\) and \(y_{2}\) into \(\left(1+x^{2}\right) y_{2}+2 x y_{1}=0\). After the substitution, this becomes \((1+x^2) \left(\frac{-2x}{(1+x^2)^2}\right) + 2x \left(\frac{1}{1+x^2}\right)\). If we simplify this, it equals zero, hence proving the equation.

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Most popular questions from this chapter

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