2. Find \( \frac{d}{dx} (2 \sin^{-1}(5x)) \) - HSC - SSCE Mathematics Extension 1 - Question 2 - 2005 - Paper 1

Using the binomial expansion:

[ (a + b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k, ] where ( a = 2x ) and ( b = -\frac{1}{x^2} ).

We want to find the term where the power of ( x ) equals zero: [ (2x)^{12-k}(-\frac{1}{x^2})^k. ] This term yields: [ 2^{12-k} (-1)^k x^{12-k-2k} = 2^{12-k} (-1)^k x^{12-3k}. ] Setting ( 12 - 3k = 0 ) gives ( k = 4 ). Thus, substituting ( k = 4 ): [ 2^{12-4}(-1)^4 = 2^8 = 256. ]

Using the product rule: [ \frac{d}{dx}(uv) = u'v + uv', ] let ( u = e^{x} ) and ( v = \cos x - 3 \sin x ).

Thus: [ u' = -\sin x - 3 \cos x, ] [ u' = e^{x}(\cos x - 3 \sin x) + e^{x}(-\sin x - 3 \cos x) = e^{x}((\cos x - 3 \sin x) - (\sin x + 3 \cos x)). ]

Using integration by parts or recognizing that we can derive it from the previous step: We can set up: [ \int e^{3x} \sin x , dx = \text{Let } I = \int e^{3x} \sin x , dx. ] Using integration by parts: [ I = -\frac{1}{2} e^{3x}(\sin x - 3 \cos x) + C.]

Starting from the differential equation: [ \frac{dT}{dt} = -k(T - 3), ] Substituting ( T = 3 + A e^{-kt} ): [ \frac{d}{dt}(3 + A e^{-kt}) = -k(3 + A e^{-kt} - 3),
\frac{d}{dt}(A e^{-kt}) = -kA e^{-kt},
] Thus verifying the equation holds. ]

Using ( T = 3 + A e^{-kt} ): [11 = 3 + A e^{-10k} \Rightarrow A = 8 + 8 e^{-10k}.
A = 8. ] Finding temperature after 15 minutes: [T(15) = 3 + 8 e^{-15k} .] ] Giving the value of temperature after 15 minutes.