Two tangents to an exponential-logarithmic curve

A curve CC is given by the equation ex2y=ln⁡x2e^{x^2}y=\ln x^2, where x≠0x\ne0.

(i) Show that 2xln⁡x2+ex2dydx=2x2x\ln x^2+e^{x^2}\frac{dy}{dx}=\frac2x. marks

(ii) Find the equation of the tangent to CC at the point PP where x=1x=1. Given that the tangent at PP meets CC again at the point QQ, find the coordinates of QQ, leaving your answer correct to 5 significant figures. marks

(iii) Determine the acute angle, in degrees, between the tangents to CC at the points PP and QQ. marks

[2022/HCI/Prelim/I/9]

Total marks: 9

Solution

(i)

Implicit differentiation with a product

A curve CC is given by ex2y=ln⁡x2e^{x^2}y=\ln x^2, x≠0x\ne0. Show that 2xln⁡x2+ex2dydx=2x2x\ln x^2+e^{x^2}\dfrac{dy}{dx}=\dfrac2x.

Hint

  1. Differentiate both sides, using the product rule on ex2ye^{x^2}y.

  2. Replace ex2ye^{x^2}y by ln⁡x2\ln x^2 using the original equation.

Answer · (i)

2xln⁡(x2)+ex2y′=2/x2x\ln(x^2)+e^{x^2}y'=2/x.

x≠0x\ne0.