How do you find the derivative of #f(x)=2/(3x^3-x^2-24x-4)#?
1 Answer
Explanation:
If you are studying maths, then you should learn the Quotient Rule for Differentiation, and practice how to use it:
# d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 # , or less formally,# (u/v)' = (v(du)-u(dv))/v^2 #
I was taught to remember the rule in word; " vdu minus udv all over v squared ". To help with the ordering I was taught to remember the acronym, VDU as in Visual Display Unit.
So with
# :. d/dx(u/v) = (v(du)/dx-u(dv)/dx)/v^2 #
# :. f'(x) = ( (3x^3-x^2-24x-4)(0) - (2)(9x^2-2x-24) ) / (3x^3-x^2-24x-4)^2#
# :. f'(x) = -(2(9x^2-2x-24) ) / (3x^3-x^2-24x-4)^2#
Incidental, we could also use the chain rule as follows;
# f(x )= 2/(3x^3-x^2-24x-4) #
# :. f(x )= 2(3x^3-x^2-24x-4)^-1 #
# :. f'(x )= 2(-1)(3x^3-x^2-24x-4)^-2 * d/dx(3x^3-x^2-24x-4) #
# :. f'(x )= -2/(3x^3-x^2-24x-4)^2 * d/dx(3x^3-x^2-24x-4) #
# :. f'(x )= -2/(3x^3-x^2-24x-4)^2 * (9x^2-2x-24) #
# :. f'(x )= -(2(9x^2-2x-24))/(3x^3-x^2-24x-4)^2 #
