Answer: (a) The approximate change is \(-4h\); (b) \(\displaystyle y-\frac{dy}{dx}-\frac13\frac{d^2y}{dx^2}=\frac{(x+1)(x-4)}{(x-3)^5}\).
Answer: (a) The approximate change is \(-4h\); (b) \(\displaystyle y-\frac{dy}{dx}-\frac13\frac{d^2y}{dx^2}=\frac{(x+1)(x-4)}{(x-3)^5}\).
(a) Write
\(y=(1+\cos^{2}x)\operatorname{cot}x.\)
Differentiate using the product rule:
\(\frac{dy}{dx}=(-2\cos x\sin x)\operatorname{cot}x -(1+\cos^{2}x)\frac{1}{\sin^{2}x}.\)
At \(x=\frac{\pi}{4}\),
\(\sin x=\cos x=\frac{\sqrt2}{2},\qquad \operatorname{cot}x=1,\qquad \sin^{2}x=\frac12.\)
Therefore
\(\frac{dy}{dx} =-2\left(\frac{\sqrt2}{2}\right)\left(\frac{\sqrt2}{2}\right)(1) -\left(1+\frac12\right)\frac{1}{1/2}.\)
So
\(\frac{dy}{dx}=-1-3=-4.\)
For a small increase \(h\), the approximate change in \(y\) is
\(\delta y\approx \frac{dy}{dx}h=-4h.\)
(b) Since
\(y=(x-3)^{-3},\)
we have
\(\frac{dy}{dx}=-3(x-3)^{-4}\)
and
\(\frac{d^2y}{dx^2}=12(x-3)^{-5}.\)
Hence
\(y-\frac{dy}{dx}-\frac13\frac{d^2y}{dx^2} =(x-3)^{-3}+3(x-3)^{-4}-4(x-3)^{-5}.\)
Put all terms over the common denominator \((x-3)^5\):
\(\frac{(x-3)^2+3(x-3)-4}{(x-3)^5}.\)
Simplify the numerator:
\((x-3)^2+3(x-3)-4=x^2-3x-4.\)
Factorise:
\(x^2-3x-4=(x+1)(x-4).\)
Therefore
\(y-\frac{dy}{dx}-\frac13\frac{d^2y}{dx^2} =\frac{(x+1)(x-4)}{(x-3)^5}.\)
