Differentiate????

Question

Differentiate with respect to x

a) [5tan4x]/4

b) In(3x-2/6x²)

Answer 1

Question a)
y = (5/4) tan (4x)
Let u = 4x
du/dx = 4
y = (5/4) tan u
dy/du = (5/4) sec ² u
dy / dx = (dy/du) x (du/dx)
dy/dx = (4)(5/4) sec² (4x)
dy/dx = 5 sec ² (4x)

Question b)
Guessing that this is meant to read as:-
y = ln [ (3x - 2) / 6x ² ]
Let u = (3x - 2) / (6x ²)
du/dx = [(6x²)(3) - (3x - 2)(12x)] / 36x^4
du/dx = [18x² - 36x² + 24x] / 36x^4
du/dx = [24x - 18x²] / 36x^4
du/dx = 6x(4 - 3x) / 36x^4
du/dx = (4 - 3x) / (6x³)
y = ln u
dy/du = 1 / u
dy/du = (6x²) (3x - 2)
dy / dx = (6x²) (3x - 2) (4 - 3x) / (6x³)
dy/dx = (3x - 2)(4 - 3x) / x

Answer 2

a) (d/dx)[5tan4x]/4= 4*5sec(4x)^2/4= 5sec(4x)^2
Use the chain rule. Also derivative of tanu is u'sec(u)^2

b) (d/dx) In(3x-2/6x²) = (3-2/3x)/(2x-1/3x²)
(d/dx)ln u = u'/u

Answer 3

a)=(5/4)(tan4x)'
=(5/4)(sec^2(4x))(4)
=5(sec^2(4x))

b)=(6x^2/3x-2)(3x-2/6x²)'
= u do the rest im bored