Find dy/dx for the following: y = x y = ln(x3x) y = (lnx)3x y = (x2 - 5x)(4 lnx)

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Explanation / Answer

1.lny=(3x-e^x)lnx =>dy/dx=y*((3x-e^x)/x-y*lnx*(3-e^x)) =x^(3x-e^x)*(3x-e^x)/x - x^(3x-e^x)lnx*(3-e^x) 2.y=3x*lnx dy/dx=3(1+lnx) 3. lny= 3x*ln(lnx) =>dy/dx=y*(3/lnx+3ln(lnx)) =(lnx)^3x(3/lnx+3ln(lnx)) 4.lny=ln(x^2-5x)+ln(4lnx)+(x-4)-ln(3x^4-1/5)-ln(5^x-4) dy/dx={(2x-5)/(x^2-5x)+1/xlnx+1-(12x^3)/(3x^4-1/5)-(5^x*ln5)/(5^x-4)}*(x^2-5x)*4lnx*e^(x-4)/((3x^4-1/5)(5^x-4)) =(2x^3-15x^2+25x+1/xlnx+1-60x^3/(15x^3-1)-5^x*ln5/(5^x-4))*(x^2-5x)*4lnx*e^(x-4)/((3x^4-1/5)(5^x-4)) 5. 2*lny=ln(x^-2-5x)+ln(lnx/ln4)+x^2+2x-ln(3-1/x)-ln(625-x) 2*dy/dx={(-2x^-3-5)/(x^-2-5x)+1/xlnx+2x-2-1/(3x^2-x)+1/(625-x)}*{((x^-2-5x)*lnx*e^(x^2+2x))/(ln4*(3-1/x)*(625-x)) little further simplification will give u the answer.......and pls rate my answer quickly

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