Exams Joint Entrance Examination Mathematics Limits, Continuity and Differentiability Question 284 Question 284 of 401 Medium MCQ
If y = ( x + 1 + x 2 ) n y=\left(x+\sqrt{1+x^2}\right)^n y = ( x + 1 + x 2 ) n , then ( 1 + x 2 ) d 2 y d x 2 + x d y d x (1+x^2)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx} ( 1 + x 2 ) d x 2 d 2 y + x d x d y is
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d y d x = n ( x + 1 + x 2 ) n − 1 ( 1 + x 1 + x 2 ) = n ( x + 1 + x 2 ) n 1 + x 2 = n y 1 + x 2 \dfrac{dy}{dx}=n\left(x+\sqrt{1+x^2}\right)^{n-1}\left(1+\dfrac{x}{\sqrt{1+x^2}}\right)=\dfrac{n\left(x+\sqrt{1+x^2}\right)^n}{\sqrt{1+x^2}}=\dfrac{ny}{\sqrt{1+x^2}} d x d y = n ( x + 1 + x 2 ) n − 1 ( 1 + 1 + x 2 x ) = 1 + x 2 n ( x + 1 + x 2 ) n = 1 + x 2 n y .
So 1 + x 2 y 1 = n y ⇒ ( 1 + x 2 ) y 1 2 = n 2 y 2 \sqrt{1+x^2}\,y_1=ny\Rightarrow(1+x^2)y_1^2=n^2y^2 1 + x 2 y 1 = n y ⇒ ( 1 + x 2 ) y 1 2 = n 2 y 2 . Differentiating: ( 1 + x 2 ) 2 y 1 y 2 + 2 x y 1 2 = n 2 ⋅ 2 y y 1 ⇒ ( 1 + x 2 ) y 2 + x y 1 = n 2 y (1+x^2)2y_1y_2+2xy_1^2=n^2\cdot2yy_1\Rightarrow(1+x^2)y_2+xy_1=n^2y ( 1 + x 2 ) 2 y 1 y 2 + 2 x y 1 2 = n 2 ⋅ 2 y y 1 ⇒ ( 1 + x 2 ) y 2 + x y 1 = n 2 y . Answer the question to unlock