Find the gradient of the curve y=2sinx/x^3 at the point x=

  1. To find the gradient of the curve we must differentiate the function. The function is in the form of a quotient, y=u/v, where u and v are functions of x. Therefore, we can use the quotient rule, dy/dx = (v (du/dx) – u (dv/dx))/ v^2. 2) We can write u = 2sinx and differentiating this we obtain du/dx = 2cosx. 3) We then take v= x^3 and differentiating this we obtain dv/dx = 3x^2 by multiplying by the power then taking one off the power (the general rule for differentiation being y=ax^n, dy/dx = anx^(n-1). 4) The quotient rule then gives, dy/dx = (v (du/dx) – u (dv/dx))/ v^2 = ( 2x^3cosx – 6x^2sinx) / x^6 = 2x^2 (xcosx – 3sinx) / x^6 = 2(xcosx – 3sinx)/x^4. 5) To find the gradient at the point Q where x=1 we substitute x=1 into dy/dx. We obtain, dy/dx = 2(cos1 – 3sin1).
HM
Answered by Holly M. Maths tutor

3533 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

The equation of curve C is 3x^2 + xy + y^2 - 4x - 6y + 7 = 0. Use implicit differentiation to find dy/dx in terms of x and y.


Prove the identity: (sinx - tanx)(cosx - cotx) = (sinx - 1)(cosx - 1)


how do integrate an equation with a surd or a fraction?


Find the equation of the tangent to the curve y = (2x -3)^3 at the point (1, - 1), giving your answer in the form y = mx + c.


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