How do you differentiate #f(x)=sqrt(ln((x+3)^4 / (x+x^2)))# using the chain rule?

1 Answer
Vishnuram L.
Jan 7, 2016

#f'(x) = 1/(2sqrt(4ln(x+3) - ln(x+x^2))) * (4/(x+3) - (1+2x)/(x+x^2))#.

Explanation:

Firstly, I'd try to simplify #f(x)# in the following way, using log laws.

#f(x) = sqrt(ln((x+3)^4) - ln(x+x^2))#

#f(x) = sqrt(4ln(x+3) - ln(x+x^2))#

#f(x) = (4ln(x+3) - ln(x+x^2))^(1/2)#

Now let's try to differentiate, the derivative of the outside with respect to the inside, which is the square root function is denoted in red, the derivative of the radicand i.e., logs is in teal. Of course, we can think of doing the chain rule inside the log functions, which I've shown in purple.

#f'(x) = color(red)((1/2) * (4ln(x+3) - ln(x+x^2))^(-1/2)) * color(teal)((color(purple)(4)/(x+3) - color(purple)((1+2x))/(x+x^2)))#.

Writing the index in radical form, we have:

#f'(x) = 1/(2sqrt(4ln(x+3) - ln(x+x^2))) * (4/(x+3) - (1+2x)/(x+x^2))#, or

#f'(x) = 1/(2sqrt(ln((x+3)^4/(x+x^2)))) * (4/(x+3) - (1+2x)/(x+x^2))#. Whichever looks more aesthetically pleasing.

Related questions
See all questions in Chain Rule
Impact of this question
1543 views around the world
You can reuse this answer
Creative Commons License