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NST First Year Maths

Some unofficial solutions to Cambridge NST IA maths papers

Paper 1

2014 Paper 1 Question 1

5th Sep 20191st Nov 2020nstmathsupervisorLeave a comment

(a)

Apply the chain rule

\frac{d}{dx}(x^2+4)^{-1} = -(x^2+4)^{-2}(2x)=-\frac{2x}{(x^2+4)^{2}}

(b)

Apply the chain rule

\frac{d}{dx}e^{\sin(x)} = e^{\sin(x)}\frac{d}{dx}\sin(x) = \cos(x)e^{\sin(x)}\

2014, Paper 1differentiation
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