Given y=x^2(1+4x)^0.5, show that dy/dx=2x(5x+1)/((1+4x)^0.5)

You have to differentiate the equation y=x^2(1+4x)^0.5 with respect to x to get dy/dx and thereby prove the result in the question. An equation composed of the product of two x terms can be differentiated with respect to x using the product rule to get dy/dx. The equation is expressed in the form y=uv, where u=x^2 and v=(1+4x)^0.5.By the product rule dy/dx=udv/dx+vdu/dx.You get dv/dx and du/dx by differentiating the u and v terms with respect to x, hence.....du/dx=2x and dv/dx=2(1+4x)^-0.5. Sub the u,v, dv/dx and du/dx terms in the product rule equation to get....dy/dx=((2x^2/(1+4x)^0.5))+(2x((1+4x)^0.5)).We now have a correct result for dy/dx but it isn't in the form desired by the question. To get dy/dx in the correct form of dy/dx=2x(5x+1)/((1+4x)^0.5), we need to firstly try and add the two terms together to get the single term.To add two fractions together you need to get the same denominator for each one, given the first term has a denominator of (1+4x)^0.5), you need to multiply the 2nd terms numerator and denominator by (1+4x)^0.5. The result of this is.....dy/dx=(2x^2)/((1+4x)^0.5))+(2x((1+4x)^0.5))*(1+4x)^0.5)/((1+4x)^0.5))The numerator of the 2nd term simplifies to 2x(1+4x), so the derivative simplifies to.....dy/dx=(2x^2)/((1+4x)^0.5))+(2x((1+4x)/((1+4x)^0.5))dy/dx=(2x^2)/((1+4x)^0.5))+(2x+8x^2)/((1+4x)^0.5))Both fractions now have the same denominator terms of ((1+4x)^0.5), so they can be added together to get ..dy/dx=(2x+10x^2)/((1+4x)^0.5))Simply this dy/dx term into the desired format by factorising the numerator and pulling out a factor of 2x.dy/dx=2x(1+5x)/((1+4x)^0.5)) Final Answer