Answer: \(\dfrac{dy}{dx}=\dfrac{2e^{2x-3}(x^2-x+1)}{(x^2+1)^2}\); \(\dfrac{dx}{dt}=\dfrac{25}{3e}\).

(a) Use the quotient rule with

\(u=e^{2x-3},\qquad v=x^2+1.\)

Then

\(u'=2e^{2x-3},\qquad v'=2x.\)

Therefore

\(\frac{dy}{dx} =\frac{(x^2+1)2e^{2x-3}-2xe^{2x-3}}{(x^2+1)^2}.\)

Factorising the numerator gives

\(\frac{dy}{dx} =\frac{2e^{2x-3}(x^2-x+1)}{(x^2+1)^2}.\)

(b) When \(x=2\),

\(\frac{dy}{dx} =\frac{2e(4-2+1)}{25} =\frac{6e}{25}.\)

Since \(y\) is increasing at \(2\) units per second,

\(\frac{dy}{dt}=2.\)

Using

\(\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt},\)

we get

\(2=\frac{6e}{25}\frac{dx}{dt}.\)

Hence

\(\frac{dx}{dt}=\frac{25}{3e}.\)