Target Mathematics

2.78: Logarithmic trigonometric differentiation

0606/23/O/N/20 — Question 4 · 9 marks

It is given that y=ln⁡(1+sin⁡x)y = \ln(1 + \sin x) for 0<x<π0 < x < \pi.
(a)
Find dydx\dfrac{\mathrm{d}y}{\mathrm{d}x}.
[2]
(b)
Find the value of dydx\dfrac{\mathrm{d}y}{\mathrm{d}x} when x=π6x = \dfrac{\pi}{6}, giving your answer in the form 1a\dfrac{1}{\sqrt{a}}, where aa is an integer.
[2]
(c)
Find the values of xx for which dydx=tan⁡x\dfrac{\mathrm{d}y}{\mathrm{d}x} = \tan x.
[5]