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To find the turning points we need to find when the differential of the equations with respect to x is equal to 0. (dy/dx = 3x² - 12 = 0) From this we find that the turning points happ...

To solve such equations we take advantage of log lawes to simplify the problem .

E.g

ln[sqrt(1-x²)] = ln[(1-x²)^1/2] = 1/2ln[1-x²]

After sim...

To solve this you need to integrate by substitution. You can spot this because the differential of the bottom of the fraction is a multiple of the top part, showing this quickly; if u = x²  + 3...

The equation we use to integrate by parts is

y = uv - ∫ v(du/dx) dx + c

so we separate dy/dx into u=(3x-4) and dv/dx=(2x²+5)

however we still need to find du/dx an...

To find the gradient of the curve at t=2 we need to find an expression for dy/dx and then substitute in for t=2. We can make use of the chain rule to find this expression because dy/dx = (dy/dt)/(dx/dt) a...