Question

(c) y = log x+2+Vx2 + 4x +1 2 (find the derivative)​

Answer 1

Answer:

The correct question is :

If y = log(x + \sqrt{x^2+1} )y=log(x+x2+1) then prove that  (x^2 +1) \frac{d^2y}{dx^2} + x \frac{dy}{dx} =0(x2+1)dx2d2y+xdxdy=0

Given the equation

y = log(x + \sqrt{x^2+1} )y=log(x+x2+1)

We have to prove that

(x^2 +1) \frac{d^2y}{dx^2} + x \frac{dy}{dx} =0(x2+1)dx2d2y+xdxdy=0

Given the equation

         y = log(x + \sqrt{x^2+1} )y=log(x+x2+1)

Differentiating with respect to 'x' , we get

         \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2+1} }(1 + \frac{2x}{2\sqrt{x^2+1} } )dxdy=x+x2+11(1+2x2+12x)

        \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2+1} }( \frac{x+\sqrt{x^2+1} }{\sqrt{x^2+1} } )dxdy=x+x2+11(x2+1x+x2+1)

        \frac{dy}{dx} = \frac{1}{\sqrt{x^2+1} }dxdy=x2+11

        Multiplying the above equation by \sqrt{x^2+1}x2+1 , we get

        \sqrt{x^2+1}\frac{dy}{dx} = 1x2+1dxdy=1

Now, Differentiating again with respect to 'x' that is the second order differential of the given function, we get

         \sqrt{x^2+1}\frac{d^2y}{dx^2} + \frac{2x}{2\sqrt{x^2+1} }\frac{dy}{dx} = 0x2+1dx2d2y+2x2+12xdxdy=0

Multiplying the above equation by \sqrt{x^2+1}x2+1 we get,

         ({x^2+1})\frac{d^2y}{dx^2} + x\frac{dy}{dx} = 0(x2+1)dx2d2y+xdxdy=0 , hence proved.

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Answer 2

Find the derivative of y = ln x^4

Find derivative cos(lnx)

1. f(x)=ln(Ax +B) Find the derivative.

Updated On: 31-1-2020

1. Find the derivative of y = ln 2