How do I differentiate (x^2 + 3x + 3)/(x+3)

In order to differentiate this fraction, you will need to use the quotient rule, which if you don't remember is:
y = u/v then dy/dx = (vdu/dx − u dv/dx)/ v^2)
First you need to identify your initial variables:
u = x^2 + 3x + 3
v = x+3
Then you need to differentiate these initial variables, which is done by multiplying the coefficient of each term by the number of the power and then subtracting one off of the power:
du/dx = 2x + 3dv/dx = 1
Remember that for number terms, the real term is 3
x^0. This means that when you multiply by the power, entire term is turned to 0.
Now that you have your variables, all you have to do is substitute them into the formula:
dy/dx = (((x+3)
(2x+3)) - ((x^2 + 3x + 3)*1))/(x + 3)^2
After you expand the brackets, this becomes
dy/dx = (2x^2 +6x + 3x + 9 - x^2 - 3x - 3)/(x+3)^2
dy/dx = (x^2 + 6x + 6)/(x+3)^2

This is the simplest form, therefore we have the answer!

Answered by Maths tutor

3439 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Find the intersection point of the line 2y=x+3 with the ellipse y^2+2x^2=3


integrate (2x^4 - 4/sqrt(x) + 3)dx


A line L is parallel to y=4x+5 and passes through the point (-1, 6). Find the equation of the line L in the form y=ax+b . Find also the coordinates of its intersections with the axes.


How to differentiate 2x^5-4x^3+x^2 with respect to x


We're here to help

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

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences