Find The Derivative Of Y = Ln Sqrt(x^2-1)/(x^2+1)

Anonymous

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Find the derivative of y = ln sqrt(x^2-1)/(x^2+1)

 

Answer No.1

MythJay

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D/( dx)(y = (log(sqrt(x^2-1)))/(x^2+1))

Answer No.2

ProGuy3

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Y = g(x) = ln [ x. sqrt( x^2 - 1 ) ] = ln x + ln [ ( x^2 - 1 )^( 1/ 2) ] = ln x + ( 1/2 ) ln ( x^2 - 1 ) ...................................... Hence g ' (x) = Dx ( ln x ) + ( 1/2 ). Dx [ ln ( x^2 - 1 ) ] = ( 1/x ) + ( 1/2 ) [ 1 / ( x^2 - 1 ) ]. Dx ( x^2 - 1 ) = ( 1/x ) + ( 1/2 ) [ 1 / ( x^2 - 1 ) ] ( 2x - 0 ) = ( 1/x ) + [ x / ( x^2 - 1 ) ] = [ 1( x^2 - 1 ) + x ( x ) ] / [ x.( x^2 - 1 ) ] = ( x^2 - 1 + x^2 ) / x ( x^2 - 1 ) = ( 2 x^2 - 1 ) / x.( x^2 - 1 )

Answer No.3

Anonymous

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Y = ln sqrt((x-1)/(x+1)) y = (1/2)ln(x - 1) - (1/2)ln(x + 1) dy/dx = (1/2)(1/(x - 1)) - (1/2)(1/(x + 1)) dy/dx = [(x + 1) - (x - 1)] / [2(x - 1)(x + 1)] dy/dx = 2 / [2(x - 1)] dy/dx = 1 / (x - 1)

Answer No.4

Anonymous

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D/dx(y(x)) = d/dx((log(sqrt(x^2-1)))/(x^2+1)) | The derivative of y(x) is y'(x): = | y'(x) = d/dx((log(sqrt(x^2-1)))/(x^2+1)) | Use the quotient rule, d/dx(u/v) = (v ( du)/( dx)-u ( dv)/( dx))/v^2, where u = log(sqrt(x^2-1)) and v = x^2+1: = | y'(x) = ((x^2+1) (d/dx(log(sqrt(x^2-1))))-log(sqrt(x^2-1)) (d/dx(x^2+1)))/(x^2+1)^2 | Use the chain rule, d/dx(log(sqrt(x^2-1))) = ( dlog(u))/( du) ( du)/( dx), where u = sqrt(x^2-1) and ( dlog(u))/( du) = 1/u: = | y'(x) = ((x^2+1) (d/dx(sqrt(x^2-1)))/sqrt(x^2-1)-log(sqrt(x^2-1)) (d/dx(x^2+1)))/(x^2+1)^2 | Use the chain rule, d/dx(sqrt(x^2-1)) = ( dsqrt(u))/( du) ( du)/( dx), where u = x^2-1 and ( dsqrt(u))/( du) = 1/(2 sqrt(u)): = | y'(x) = (((x^2+1) (d/dx(x^2-1))/(2 sqrt(x^2-1)))/sqrt(x^2-1)-log(sqrt(x^2-1)) (d/dx(x^2+1)))/(x^2+1)^2 | Differentiate the sum term by term: = | y'(x) = (((x^2+1) (d/dx(x^2)+d/dx(-1)))/(2 (x^2-1))-log(sqrt(x^2-1)) (d/dx(x^2+1)))/(x^2+1)^2 | The derivative of -1 is zero: = | y'(x) = (((x^2+1) (d/dx(x^2)+0))/(2 (x^2-1))-log(sqrt(x^2-1)) (d/dx(x^2+1)))/(x^2+1)^2 | The derivative of x^2 is 2 x: = | y'(x) = (((x^2+1) (2 x))/(2 (x^2-1))-log(sqrt(x^2-1)) (d/dx(x^2+1)))/(x^2+1)^2 | Differentiate the sum term by term: = | y'(x) = ((x (x^2+1))/(x^2-1)-log(sqrt(x^2-1)) (d/dx(x^2)+d/dx(1)))/(x^2+1)^2 | The derivative of 1 is zero: = | y'(x) = ((x (x^2+1))/(x^2-1)-log(sqrt(x^2-1)) (d/dx(x^2)+0))/(x^2+1)^2 | The derivative of x^2 is 2 x: = | y'(x) = ((x (x^2+1))/(x^2-1)-(2 x) log(sqrt(x^2-1)))/(x^2+1)^2

Answer No.5

TinyAsian

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Votes: 81
Y'= [(x2 +1)/(√(x2 -1))(x)/(x2 -1)3/2 - ln(√(x2 -1)(2x)]/[x2 + 1]2

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