How do you use the chain rule to differentiate #1/ln(4x)#?

Question

Samantha Adkison · Accepted Answer

Let #u = 4x#, #v = ln(u) = ln(4x)#. Differentiating the above equation yields #frac{du}{dx} = 4#
#frac{dv}{du} = 1/u# Then,  #1/ln(4x) = 1/v# #frac{d}{dx}(1/ln(4x)) = frac{d}{dx}(1/v)# #= frac{d}{dv}(1/v) frac{dv}{du} frac{du}{dx}# #= (-1/v^2) * (1/u) * (4)# #= (-1/(ln(4x))^2) * (1/{4x}) * (4)# #= -1/(x (ln(4x))^2 )#

Charles Anctil · Answer

undefined To differentiate $ \frac{1}{\ln(4x)} $ using the chain rule, follow these steps:

1. Identify the inner function, which is $ \ln(4x) $.
2. Differentiate the inner function with respect to $ x $, which yields $ \frac{d}{dx}(\ln(4x)) = \frac{1}{4x} \cdot 4 = \frac{4}{4x} = \frac{1}{x} $.
3. Rewrite the original function as $ y = \frac{1}{u} $, where $ u = \ln(4x) $.
4. Apply the chain rule: $ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} $.
5. Substitute the derivatives: $ \frac{dy}{dx} = \frac{d}{du}\left(\frac{1}{u}\right) \cdot \frac{1}{x} $.
6. Differentiate $ \frac{1}{u} $ with respect to $ u $, yielding $ -\frac{1}{u^2} $.
7. Combine the results: $ \frac{dy}{dx} = -\frac{1}{(\ln(4x))^2} \cdot \frac{1}{x} $.
8. Simplify the expression if necessary.

So, the derivative of $ \frac{1}{\ln(4x)} $ with respect to $ x $ is $ -\frac{1}{(\ln(4x))^2} \cdot \frac{1}{x} $.