# Find the derivative of  function: `1) y=lnx/(2x-1) ` `2) y=e^x*(x^2-2x+2) ` `3) y= (x+3)^3/(xsqrtx)` ``

Asked on by lienetom

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

`1. y= lnx/(2x-1) `

`==gt (dy)/(dx) = ((lnx)'(2x-1) - (lnx)(2x-1)')/(2x-1)^2 `

`==gt (dy)/(dx) = ((1/x)(2x-1) - (lnx)(2))/(2x-1)^2 `

`==gt (dy)/(dx) = (2- 1/x - 2lnx)/(2x-1)^2 `

`==gt (dy)/(dx)= (2x-2xlnx -1)/(x(2x-1)^2)`

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`2. y= e^x (x^2 - 2x+2) `

`==gt (dy)/(dx) = (e^x)' (x^2-2x+1) + (e^x)(x^2-2x+2)' `

`==gt (dy)/(dx) = e^x (x^2-2x+1) + e^x (2x-2) `

`==gt (dy)/(dx) = (x^2)e^x - 2xe^x + e^x + xe^2x - 2e^x == (dy)/(dx) = x^2 e^x - xe^x - e^x `

`==gt (dy)/(dx) = e^x (x^2 -x -1)`

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`3. y= (x+3)^2 /(xsqrtx) `

`==gt (dy)/(dx) = (((x+3)^3)'(xsqrtx) - (x+3)^3 (xsqrtx)')/(xsqrtx)^2 `

`==gt (dy)/(dx) = (3(x+3)^2 xsqrtx - (x+3)^3 (-sqrtx/2 +sqrtx))/x^3 `

`==gt (dy)/(dx) = (3(x+3)^2 xsqrtx +((x+3)^3 sqrtx)/2 )/x^3`

`==gt (dy)/(dx) = (6(x+3)^2 xsqrtx + (x+3)^3 sqrtx)/(2x^3)`

`==gt (dy)/(dx) = ((x+3)^2 sqrtx (7x+3))/(2x^3)`

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