One problem I've been asked to solve is giving me some trouble on the particular sequence to solve. Find the differential of

$$g(t)=\frac{10 \log_{4}t}{t}$$

Looking at the problem, you can see that the 10 is a constant, and can be pulled out of the function and that the quotient rule applies

$$g'(t)= 10[\frac{\log_{4}t}{t}]$$ $$g'(t)=10[\frac{\log_{4}t[\frac{d}{dx}(t)]-t[\frac{d}{dx}\log_4t]}{t^2}]$$ $$g'(t)= 10[\frac{\log_4t-t(\frac{1}{\ln(4)t})}{t^2}]$$ $$g'(t)= 10[\frac{\log_4t-\frac{t}{\ln(4)t}}{t^2}]$$ $$g'(t)=10[\frac{t^2}{\log_4- 2t \ln(4)}]$$

I got stuck right there, since I'm not sure my last line works, so I took a look at the book solution, which is shown below:

$$g(t)= \frac{10 \log_4t}{t} = \frac{10}{\ln 4}\cdot\frac{\ln(t)}{t}$$

So my questions are:

$\cdot$There are 3 elements of the numerator, so how does the $\log_4$ morph into the denominator of $\frac{10}{\ln 4}$? $\cdot$The book example shows that the quotient differentiation is done only on $\frac{\ln(t)}{t}$, but why not on the other fraction? Is it due to those values being constants and not open to differentiation?
$\cdot$Is the last line of my own work accurate?

  • 1
    $\begingroup$ Let $z=\log_4 t$. Then $4^z=t$. Take the $\ln$ of both sides. We get $z\ln 4=\ln t$. So $\log_4 t=\ln t/\ln 4$. (This may be also found from a formula you once memorized.) So you want to differentiate $10\ln t/(t\ln 4)$. This should not give you trouble. $\endgroup$ – André Nicolas Jul 17 '11 at 21:48
  • $\begingroup$ About your various questions, there are some errors, including algebra errors. You quoted the derivative of $\log_4 t$ correctly, presumably from a list of formulas. I would not be able to remember such a thing, so I would always derive it as in the comment above. $\endgroup$ – André Nicolas Jul 17 '11 at 22:22

The expression $\log_4(t)$ is not multiplication. It is the application of a function. Your mystery will be revealed by the change of base formula, $$\log_4(t) = {\log(t)\over \log(4)}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.