Differentiate sin(x^3) with respect to y

For this we must use the chain rule. We start by defining x3 as a new variable, u = x3 Can then rewrite the expression as y = sin(u) Chain rule tells us that dy/dx = (dy/du)(du/dx) We can calculate these individidually. dy/du = cos(u)  du/dx = 3x2 Finally we can then say, dy/dx = dy/du * du/dx = cos(u) * 3x2 = 3x2cos(x3)

Answered by Lloyd B. Maths tutor

5875 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Using the trigonometric identity for tan(A + B), prove that tan(3x)=(3tan(x)-tan^3(x))/(1-3tan^2(x))


When do you know to use integration by parts?


Using the binomial theorem, find the coefficient of x^4*y^5 in (x-2y)^9.


Express [1+4(square root)7] /[ 5+ 2(square root)7] in the form m + n (square root)7 , where m and n are integers.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences