2.27: Exponential product: stationary point, tangent and integral

0606/21/O/N/19 — Question 8 · 10 marks

The equation of a curve is given by y=xe−2xy = x\mathrm{e}^{-2x}.
(i)
Find dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}.
[3]
(ii)
Find the exact coordinates of the stationary point on the curve y=xe−2xy = x\mathrm{e}^{-2x}.
[2]
(iii)
Find, in terms of e\mathrm{e}, the equation of the tangent to the curve y=xe−2xy = x\mathrm{e}^{-2x} at the point (1,1e2)\left(1, \dfrac{1}{\mathrm{e}^{2}}\right).
[2]
(iv)
Using your answer to part (i), find ∫xe−2x dx\displaystyle\int x\mathrm{e}^{-2x}\,\mathrm{d}x.
[3]