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In this question, we have the product of two separate terms, so we will choose to use the product rule for this question. Recall, for f(x) = u(x) v(x): f'(x) = u'(x) v(x) + u(x) v'(x). Here, we can set u(x) ...

At a stationary point, the gradient/slope of the graph is 0. To find the gradient of y, we differentiate with respect to x.This gives us dy/dx = 2x + 1. Since we want to find where the gradient is 0, we set ...

This is a standard integral of the type f'(x)*f(x) n . To find the solution, we trial d/dx(f(x) n+1 ). d/dx(sin 5 (x)) = 5sin 4 (x)cos(x). this looks similar to the integral we were asked to solve, apart fro...

You would first of all establish which differentiation rule is required, for this question it would be useful to use the product rule splitting xcos(x) into x multiplied by cos(x). We can label u = x and v =...

The vector from (1,-1,2) to (2,2,2) will have direction (2,2,2)-(1,-1,2) which is (1,3,0) so the vector equation is r =(1,-1,2) + s(1,3,0) where s is a scalar.