Logarithmic Differentiation Simplifying

I've been given this problem and I've been working on it for a week on and off trying to figure out how to simplify it so it looks a bit cleaner. I can get the derivative without issue (I hope it's right) but my simplifying bag-of-tricks is not as developed as it should be.

Original question-- Use logarithmic differentiation to find the derivative:

$\displaystyle y=\frac{3^{-5x}(2x-3)^\frac{4}{3}}{(7x-5)^\frac{2}{5}(3x-7)^\frac{3}{4}}$

$\displaystyle y'=\frac{3^{-5x}(2x-3)^\frac{4}{3}}{(7x-5)^\frac{2}{5}(3x-7)^\frac{3}{4}}[-5ln(3)+\frac{8}{3(2x-3)}-\frac{14}{5(7x-5)}-\frac{9}{4(3x-7)}}]$

Is my instructor just trying to make me go mad or am I missing something? Trying to learn when to stop simplifying is difficult at times.

Re: Logarithmic Differentiation Simplifying

Hello, Maskawisewin!

Quote:

I've been given this problem and I've been working on it for a week on and off

trying to figure out how to simplify it so it looks a bit cleaner.

I can get the derivative without issue (I hope it's right),

but my simplifying bag-of-tricks is not as developed as it should be.

Original question -- Use logarithmic differentiation to find the derivative:

. . $\displaystyle y\:=\:\frac{3^{\text{-}5x}(2x-3)^\frac{4}{3}}{(7x-5)^\frac{2}{5}(3x-7)^\frac{3}{4}}$

$\displaystyle y'\:=\:\frac{3^{\text{-}5x}(2x-3)^\frac{4}{3}}{(7x-5)^\frac{2}{5}(3x-7)^\frac{3}{4}}\left[\text{-}5\ln(3)+\frac{8}{3(2x-3)}-\frac{14}{5(7x-5)}-\frac{9}{4(3x-7)}}\right]$

Is my instructor just trying to make me go mad or am I missing something?

. . It depends . . . Is that your answer or the book's answer?

Trying to learn when to stop simplifying is difficult at times.

If that is *your* answer, good work!

. . I got the same answer.

If that is the book's answer (and I assume you didn't get it),

. . I need to see your work to make any comments.

Re: Logarithmic Differentiation Simplifying

No that's my answer. I like to double check my work when I think I'm wrong. Thank you for the confidence.

Re: Logarithmic Differentiation Simplifying

If you like to double check your WORK, then you need to POST your work, not just the final answer...

Re: Logarithmic Differentiation Simplifying

The original post requests if this question can be simplified any further. I had the question up to that point, which turned out to be the final answer. I actually thought I was posting my work because I did not have the confidence to say "it's done". You told me "it's done" and I breathed a sigh of relief because I was going crazy trying to simplify the expression further.

Edit: Thought the user Prove It was Soroban. Apologies Soroban.

Re: Logarithmic Differentiation Simplifying

Maskawisewin, you could try putting everything under a common denominator. That would be the only realistic "simplification" you can expect from a complicated expression. That said, if you put everything under a common denominator you will get the same result as if you just used the quotient and product rule to find the derivative.

As far as the expression itself, it's perfectly fine as it is.