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Solution — 2017 Question 2 Differentiation & Applications| Integration & Applications

(a)
y=15x−2=(5x−2)−12dydx=−12(5x−2)−32(5)=−52(5x−2)−32=−52(5x−2)32∎
(b)
y=(2x2−1)2x2=4x4−4x2+1x2=4x2−4+1x2=4x2−4+x−2
∫(2x2−1)2x2dx=∫(4x2−4+x−2)dx=4(x33(1))−4x+x−1−(1)+c=4x33−4x−1x+c∎
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