use logarithmic differentiation to find dy/dx

y = (5x-3)/(3x+2)^3

Question

use logarithmic differentiation to find dy/dx

y = (5x-3)/(3x+2)^3

konradzuse · Answer

Apparently from my last question like this I did normal differentiation...

anonymous · Answer

not sure what logarithmic differentiation is ... it's been a while lol let me check

konradzuse · Answer

http://www.wolframalpha.com/input/?i=logarithmic+differentiation++x+sqrt%28x2+%2B+9+%29

anonymous · Answer

take the ln of boht sides

precal · Answer

why not just use the quotient rule?

anonymous · Answer

since your doing logarithmic differentiation, yes, @Outkast3r09 , is correct, you'll need to take the ln of both sides...

anonymous · Answer

the question just asks to do it that way for some reason but whatever


you should get

$$ln(y)=ln(5x-3)-ln((3x+2)^3)$$ used the ln(a/b)=ln(a)-ln(b)

anonymous · Answer

differentiate both side in respects to x

precal · Answer

|dw:1340671460430:dw|

konradzuse · Answer

so ln(y) = 5/(5x-3) - (3(3x+2)^2)/(3x+2)^3) *3?

konradzuse · Answer

?

anonymous · Answer

gotta do both sides the left side is once again chain rule

where u =ln(y)

$$\frac{du}{dx}=\frac{du}{dy}*\frac{dy}{dx}$$

anonymous · Answer

the right side is correct though

konradzuse · Answer

so 1/y * null?

konradzuse · Answer

since y' of y would be 1 actually...

konradzuse · Answer

not null

precal · Answer

|dw:1340672041404:dw|