Given y = 2sin(θ) and x = 3cos(θ) find dy/dx.

The function is defined parametrically so we usually approach these questions using chain rule.Recall that: dy/dθ * dθ/dx = dy/dx So we will need to differentiate each expression individually then multiply them together.Differentiating the first with respect to θ we get:(1)   dy/dθ = 2cos(θ) ,then the expression for x gives us: dx/dθ = -3sin(θ) , We can then remember that differentials behave as fractions so we can flip both sides to get:(2)  dθ/dx = -1/3sin(θ) . Remembering chain rule we can multiply (1)*(2) to get dy/dx: dy/dθ * dθ/dx = 2cos(θ) * -1/3sin(θ) --> dy/dx = -2cos(θ)/3sin(θ)

JC
Answered by Jacob C. Maths tutor

5467 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

What is the velocity of the line from vector A(3i+2j+5k) to vector B(10i-3j+2k)?


Given that y = ((4x+3)^5)(sin2x), find dy/dx


Integrate xsin(x) with respect to x


Let f(x) = 3x^4 - 8x^3 - 3. Find the x- values of the stationary points of this function.


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