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.

ML

Related Maths A Level answers

All answers ▸

Given y(x+y)=3 evaluate dy/dx when y=1


Find the equation for the tangent to the curve y^3 + x^3 + 3x^2 + 2y + 8 = 0 at the point (2,1)


A curve (C) with equation y=3x^(0.5)-x^(1.5) cuts the X axis at point A and the origin, calculate the co-ordinates of point A.


integrate 5x + 3(square root of x)