How do you find the derivative of #log _ 3 (2x) / x^2#?
1 Answer
Explanation:
First, rewrite the logarithm in terms of the natural logarithm, since we know how to differentiate the natural logarithm.
It can be rewritten using the change of base formula:
Thus, the function can be written as
#f(x)=log_3(2x)/x^2=1/x^2ln(2x)/ln(3)=ln(2x)/(x^2ln(3))#
To differentiate this, use the quotient rule.
#f'(x)=(x^2ln(3)d/dx[ln(2x)]-ln(2x)d/dx[x^2ln(3)])/(x^2ln(3))^2#
We can find each of these derivatives individually:
The first requires the chain rule:
#d/dx[ln(2x)]=1/(2x)*d/dx[2x]=2/(2x)=1/x#
This can also be found through first splitting up
#d/dx[ln(2)+ln(x)]=d/dx[ln(x)]=1/x#
To find the second, just use the power rule and remember that
#d/dx[x^2ln(3)]=2xln(3)#
Plugging both these back in yields
#f'(x)=(x^2ln(3)(1/x)-ln(2x)(2xln(3)))/(x^4ln^2(3))#
Simplify.
#f'(x)=(xln(3)-2xln(2x)ln(3))/(x^4ln^2(3))#
#f'(x)=(xln(3)(1-2ln(2x)))/(x^4ln^2(3))#
#f'(x)=(1-2ln(2x))/(x^3ln(3))#