# How do you find the derivative of #y = ln[(x+3)/x^2)]#?

##### 2 Answers

#### Explanation:

We let

Then

The chain rule states that

#dy/dx = 1/u xx -(x + 6)/x^3#

#dy/dx= (-(x + 6)/x^3)/((x + 3)/x^2)#

#dy/dx= -(x + 6)/x^3 xx x^2/(x + 3)#

#dy/dx = -(x +6)/(x(x +3))#

Hopefully this helps!

We can also rewrite this the logarithm rules

#y=ln((x+3)/x^2)=ln(x+3)-ln(x^2)#

#color(white)(y=ln((x+3)/x^2))=ln(x+3)-2ln(x)#

We should know that

Then:

#dy/dx=1/(x+3)*d/dx(x+3)-2(1/x)#

Since

#dy/dx=1/(x+3)-2/x#

Finding a common denominator:

#dy/dx=(x-2(x+3))/(x(x+3))#

#color(white)(dy/dx)=(-x-6)/(x(x+3))#

#color(white)(dy/dx)=-(x+6)/(x(x+3))#