The curve C has an equation y = sin(2x)cos(x)^2. Find dy/dx. Find normal to curve at x = pi/3 rad, giving answer in exact form.

Student should use a combination of trigonometric identities, product rule and chain rule to find dy/dx.This can be done by applying product rule, obtainingdy/dx = sin(2x). d[cos(x)^2]/dx + cos(x)^2. d[sin(2x)]/dxthen using the chain rule to find d[cos(x)^2]/dx = -2cos(x)sin(x) andd[sin(2x]/dx = 2cos(2x).Alternatively, the student can rewrite cos(x)^2 as ½[cos(2x)-1] then differentiate in this form.Student should finddy/dx = 2cos(x)[cos(x)cos(2x) – sin(x)sin(2x)].
Substituting x= pi/3 into equation of curve returns corresponding y-value at that point, and substituting x = pi/3 into dy/dx returns gradient of tangent at that point. Diving -1 by this returns the gradient of the normal at that point. Using equation of a straight line y=mx+b, student should obtain b by rearranging at substituting x and y values found earlier. Students should find y = x + [(3sqrt(3) – 8pi)/24)]

SC
Answered by Sunchi C. Maths tutor

4157 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Parlami di cosa hai fatto durante le vacanze di Natale.


The gradient of the curve at point (x,y) is given by dy/dx = [7 sqrt(x^5)] -4. where x>0. Find the equation of the curve given that the curve passes through the point 1,3.


How does integration by parts work ad when to use it?


Represent in partial fraction form the expression x/x^2-3x+2


We're here to help

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

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning