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# Differentiate the following with respect to x and obtain `dy/dx` . `y = x^2 +sqrt(x^2 +...

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Differentiate the following with respect to x and obtain `dy/dx` .

`y = x^2 +sqrt(x^2 + sqrt (x^2 +.......oo))` .

Posted by user8235304 on July 4, 2013 at 5:41 PM via web and tagged with calculus, math

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The function `y = x^2 + sqrt(x^2 + sqrt(x^2 + sqrt(x^2+ ...oo)))` .

The series `sqrt(x^2 + sqrt(x^2 + sqrt(x^2+ ...oo)))` goes on for an infinite number of terms.

`sqrt y = sqrt(x^2 + sqrt(x^2 + sqrt(x^2 + sqrt(x^2+ ...oo))))`

This gives `y = x^2 + sqrt y`

Taking the derivative with respect to x of both the sides.

`dy/dx = 2x + (1/2)*(1/sqrt y)*(dy/dx)`

=> `(dy/dx)(1 - 1/(2*sqrt y)) = 2x`

=> `(dy/dx)((2*sqrt y - 1)/(2*sqrt y)) = 2x`

=>  `(dy/dx)((2*sqrt y - 1)/(2*sqrt y)) = 2x `

But `sqrt y = y - x^2`

=> `dy/dx = (4x*(y - x^2))/(2*(y - x^2) - 1) `

The required derivative is `dy/dx = (4x*(y - x^2))/(2*(y - x^2) - 1) `

Posted by justaguide on July 4, 2013 at 5:56 PM (Answer #1)