How would you differentiate ln(sin(3x))?

To answer this question we require the chain rule, which states that dy/dx=(dy/du)*(du/dx)

To use this formula in our question, we can let y=ln(sin(3x))=ln(u) where u=sin(3x)

Firstly, using a standard result we have dy/du=1/u

Secondly, we must work out du/dx. Another standard result is that d/dx(sin(ax))=acos(ax) for any constant number a. This means du/dx=3cos(3x)

Putting the two parts together, we find that our answer, given by dy/dx, is equal to

3*cos(3x)1/u=(3cos(3x))/(ln(sin(3x)))

HD
Answered by Hannah D. Maths tutor

11089 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How would I find the indefinite integral of x*cos(x) dx


Find the value of (cos(x) + sec(x))^2 with respect to x when evauated between pi/4 and 0


A curve has the equation y=12+3x^4. Find dy/dx.


What actually are sin, cos and tan?


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences