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

3936 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the equation of the straight line tangent to the curve y=2x^3+3x^2-4x+7, at the point x=-2.


y=x^2, find dy/dx


Factorise completely ( x − 4x^3)


How would I sketch the graph sin(x) + sin(2x - π/2) in my exam?


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:

© 2025 by IXL Learning