Solve the question.

Question

 y \(=\frac{x(1-x^2)^2}{x(1+x^2)^{1/2}}\)

Answer 1

(a) y \(=\frac{x(1-x^2)^2}{x(1+x^2)^{1/2}}=\frac{(1-x^2)^2}{(1+x^2)^{1/2}}\)

⇒ ln y = 2 ln(1-x²) - 1/2 ln(1+x²)

Differentiate w.r.t. x, we get

1/y \(\frac{dy}{dx}=\frac{2}{1-x^2}\) x -2x \(-\frac{1}{2(1+x^2)}\) x 2x (using chain rule)

⇒ \(\frac{dy}{dx}=y\left(\frac{-4x}{1-x^2}-\frac{x}{1+x^2}\right)\)

\(=y\left(\frac{-x(4+4x^2+1-x^2)}{(1-x^2)(1+x^2)}\right)\)

\(=\frac{(1-x^2)^2}{(1+x^2)^{1/2}}\times\frac{-x(3x^2+5)}{(1-x^2)(1+x^2)}\)

\(=\frac{-x(1-x^2)(3x^2+5)}{(1+x^2)^{3/2}}\)

(b) (i) y \(=ln(x+\sqrt{1+x^2})\)

⇒ e^y = x+\(\sqrt{1+x^2}\) (by taking anti log)

⇒ e^y \(\frac{dy}{dx}=1+\frac{1}{2\sqrt{1+x^2}}\) x 2x (By differentiating w.r.t. x)

⇒ e^y \(\frac{dy}{dx}=1+\frac{x}{\sqrt{1+x^2}}\) 

⇒ e^y \(\left(\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2\right)\) \(=\cfrac{\sqrt{1+x^2}\times1-x\times\frac{1}{2\sqrt{1+x^2}}\times2x}{(1+x^2)}\) (By differentiating w.r.t. x)

⇒ e^y \(\left(\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2\right)\) \(=\frac{1+x^2-x^2}{(1+x^2)^{3/2}}\)

⇒ \(\frac{d^2y}{dx^2}+(\frac{dy}{dx})^2\) \(=\frac{1}{(1+x^2)^{3/2}}\times\frac{1}{e^y}\)

⇒ \(\frac{d^2y}{dx^2}\) \(=\frac{1}{(x+\sqrt{1+x^2})(1+x^2)^{3/2}}-(\frac{dy}{dx})^2\)

\(=\frac{1}{(x+\sqrt{1+x^2})(1+x^2)^{3/2}}\) \(-\cfrac{(1+\frac{x}{\sqrt{1+x^2}})^2}{(x+\sqrt{1+x^2})^2}\)

\(=\frac{1}{(x+\sqrt{1+x^2})(1+x^2)^{3/2}}\) \(-\cfrac{(x+\sqrt{1+x^2})^2}{(x+\sqrt{1+x^2})^2(1+x^2)}\)

\(=\frac{1}{(1+x^2)^{3/2}(x+\sqrt{1+x^2})}-\frac{1}{1+x^2}\)

(ii) y = ln(ln tan x)

⇒ e^y = ln (tan x)

⇒ e^y \(\frac{dy}{dx}=\frac{1}{tan\,x}\times sec^2x\) \(=\frac{cos\,x}{sin\,x}\times\frac{1}{cos^2x}\)

\(=\frac{2}{2\,sin\,x\,cos\,x}=\frac{2}{sin\,2x}\) = 2 cosec 2x

⇒ \(\frac{dy}{dx}=\frac{2\,coses\,2x}{e^y}=\frac{2\,coses\,2x}{ln(tan\,x)}\)

And \(\left(e^y\left(\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\right)\right)\) = -4 coses 2x cot 2x

⇒ \(\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\) \(=\frac{-4\,coses\,2x\,cot\,2x}{ln\,tan\,x}\) (∵ e^y = ln tan x)

⇒ \(\frac{d^2y}{dx^2}\) \(=\frac{-4\,coses\,2x\,cot\,2x}{ln\,tan\,x}-\frac{4\,coses^22x}{(ln(tan\,x))^2}\)