Find dy/dx of the equation y=x^2 ln⁡(2x^2+1).

This question requires the use of two key rules of differentiating. The first of these is the product rule. The product rule describes a way of differentiating an equation which is the product of two different parts; which are here x^2 and ln⁡(2x^2+1). The product rule is d/dx (v.u)= du/dx . v +dv/dx . u, where u and v are the two different parts of the equation. In more simple terms, you differentiate the first aspect, and times by the second, then add the differential of the second multiplied by the first. So x^2 will differentiate to 2x, and the first part of our equation is this multiplied by the second term: => 2x[ln⁡(2x^2+1)] For the second part we must differentiate the logarithmic second term. The rule for doing this is that you must differentiate the term in the logarithm, and divide this by itself. ln⁡(2x^2+1) then will differentiate to 4x/(2x^2+1). To complete the second term, we must multiply by the initial. => [4x .(x^2)]/ (2x^2+1) Now it is just a case of summing the two terms and tidying up to get our answer, dy/dx= 2x[ln⁡(2x^2+1)] + (4x^3)/(2x^2+1)