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

Using the binomial theorem, we expand:

[ (a + b)^{n} = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^{k} ]

Here, ( a = 2x ), ( b = -\frac{1}{x^2} ), and ( n = 12 ). We look for the value of ( k ) such that the total power of ( x ) equals zero:

[ (2x)^{12-k} \left(-\frac{1}{x^2}\right)^{k} \implies 2^{12-k} x^{12-k} (-1)^{k} x^{-2k} = 2^{12-k}(-1)^{k}x^{12 -3k} ]

Setting ( 12 - 3k = 0 ) gives ( k = 4 ). Substituting ( k ) back:

[ 2^{12-4} (-1)^{4} = 2^{8} = 256 ]

Using the product rule, we differentiate:

[ \frac{d}{dx} (e^{x}(\cos x - 3\sin x)) = 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) ]

This simplifies to:

[ e^{x}(-2\sin x - 2\cos x) ]

To find ( \int e^{3x}\sin x,dx ), we can use integration by parts or the method of undetermined coefficients, leading to:

Let ( I = \int e^{3x}\sin x ,dx ).

Setting up the integration by parts method twice will yield:

[ I = \frac{1}{10}(\sin x - 3\cos x)e^{3x} + C ]

To show ( T = 3 + A e^{-kt} ) satisfies ( \frac{dT}{dt} = -k(T - 3) ):

First, differentiate:

[ \frac{dT}{dt} = -kA e^{-kt} ]

Substituting ( T ):

[ -k((3 + A e^{-kt}) - 3) = -k Ae^{-kt} = -kAe^{-kt} ]

Thus, the equation holds true.

Given the initial temperature is 25°C, we can substitute into:

[ T(10) = 3 + A e^{-10k} = 11 \implies A e^{-10k} = 8 \implies A = 8 e^{10k} ]

Now to find ( T(15) ):

[ T(15) = 3 + 8 e^{10k}e^{-15k} = 3 + 8 e^{-5k}]

Substituting the calculated ( A ) gives the final temperature after 15 minutes.