Differentiate y = xe^(2x).

We want to find dy/dx. We find this using the product rule by setting the functions f(x) = x and g(x) = e2x. With these functions, we can write the equation as y = f(x)g(x), so by applying the product rule, we have that dy/dx = f'(x)g(x) + f(x)g'(x). To calculate g'(x), we use the chain rule.If we write h(x) = 2x, then g(x) = e2x = eh(x). So by using the chain rule and the fact that ex differentiates to itself, we have that g'(x) = h'(x)eh(x) = 2e2x. Therefore by going back to the equation which we found by the product rule, dy/dx = f'(x)g(x) + f(x)g'(x) = (1)(e2x) + (x)(2e2x) = e2x + 2xe2x. We can factorise this to get dy/dx = (1 + 2x)e2x.

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Answered by Matthew L. Maths tutor

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