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We'll need to use the product rule.

Let's take u = x^2 -> du/dx = 2x, and v = ln(x) -> dv/dx = 1/x

Then dy/dx = x^2(1/x) + 2xln(x) = x + 2xln(x)

Substituting our x value gives (...

5w -3 = 3w + 15 

Need to get the numbers on one side, and letters on the other side.

5w = 3w + 15 + 3 

5w - 3w = 15 + 3

2w = 18

w = 18/2

w = 9

A) We need to find dy/dx. We know that dy/dx =(dy/dt)/(dx/dt). Differentiating each parametric equation we get: dx/dt= -2sin(2t) {using the chain rule and knowing that cos differentiates to -sin.} dy/dt= ...

Answer is 8.30am Using the speed = distance/time reference the student would do distance/speed or 5/10 = 0.5 hours