How do I differentiate y = ln(sin(3x))?

So we initially have the relationship y = ln(sin(3x)). As the left hand side is a function of the variable y and the right hand side is a function of the variable x i.e they are implicitly related, we need to use implicit differentiation. This means we differentiate each side of the equation seperately with respect to the functional variable, in this case x. Now differentiate the LHS, which is simple enough

d/dx(y) = dy/dx = f'(x)

To differentiate the RHS, we note that the differential of a logarithm ln(g(x)) is given by g'(x) / g(x), a relationship that can be found easily by integrating. So in this example we have that g(x) = sin(3x), therefore we need to use the chain rule to differentiate it. The differential of sin(3x) is thus cos(3x) multiplied by the differential of 3x which is 3, therefore g'(x) = 3cos(3x)  and so the differential of the RHS is 3cos(3x)/sin(3x) = 3cot(3x)

We thus have dy/dx = 3cot(3x)

IW
Answered by Ifan W. Maths tutor

18445 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Given that: 2tanθsinθ = 4 - 3cosθ , show that: 0 = cos²θ - 4cosθ + 2 .


The function f(x) is defined by f(x) = 1 + 2 sin (3x), − π/ 6 ≤ x ≤ π/ 6 . You are given that this function has an inverse, f^ −1 (x). Find f^ −1 (x) and its domain


Express the following in partial fractions: (x^2+4x+10)/(x+3)(x+4)(x+5)


Is a line ax+by+c=0 tangent to a circle?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences