Given f(x) = (x^(2)+(3*x)+1)/(x^(2)+(5*x)+8), find f'(x) and simplify your answer.

Taking: f(x) = (x^(2)+(3x)+1)/(x^(2)+(5x)+8)
An application of the quotient rule of differentiation is required. This rule is given as:
Where g(x) = u(x)/v(x), g'(x) = ((u'(x)v(x))-(v'(x)u(x)))/(v(x)^(2))
Hence for the case of the f(x) given, f(x) = (x^(2)+(3
x)+1)/(x^(2)+(5
x)+8)is deconstructed to:
u(x) = x^(2)+(3x)+1v(x)=x^(2)+(5x)+8
Hence:
(v(x))^2 = (x^(2)+(5x)+8)^2
Applying a simple method of differentiation gives:
u'(x)= (2
x)+3v'(x)= (2x)+5
Thus, bringing all the constituents together and entering them into the quotient rule formula:
f'(x) = ((u'(x)v(x))-(v'(x)u(x)))/(v(x)^(2))f'(x)= ((((2x)+3)(x^(2)+(5
x)+8))-(((2x)+5)x^(2)+(3x)+1))/(x^(2)+(5x)+8)^2)
Expanding and collecting like terms:
f'(x)=(2x^(2)+(14x)+19)/((x^(2)+(3*x)+1)^2)
This is as far as this expression can be simplified and hence the question has been answered fully.
Teachable points:Difference in degree is clearly -1 as is required by the definition of differentiation from first principlesThe f'(x) can be described as the rate of change of f(x) and can be used to quantify how f(x) varies as x variesFurther investigation into the graph of y=f(x) could occur from this, eventually allowing the plotting of this graph

Answered by Andrew M. Maths tutor

1413 Views

See similar Maths Scottish Highers tutors

Related Maths Scottish Highers answers

All answers ▸

Find the stationery points of x^3 + 3x^2 - 24x + 7 and determine whether the slope is increasing or decreasing at x=3.


show y=3x-5 is tangent to x^2 + y^2 +2x -4y - 5 = 0 and the point where they touch


Determine for what values of c, f(x)=4x^2-(2c+8)x+4 has no real roots.


A triangle has vertices A(-3,5), B(7,9) and C(2,11). What is the equation of the median that passes through the vertex 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