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I've tried to differentiate the following function:

$$f(t)=\frac{te^{\tan (t)}}{ln(3t+1)}$$

But I am confused at what I should do (and perhaps I forgot some identities too), I've learned the rudiments of differential calculus, I've learned about the sum/product/division of derivatives and the chain rule. My understanding of these operations is that they are tools that allow one to decompose complicated derivatives in various pieces so one can derive the simplest derivatives one by one.

I've tried the following: Using the chain rule, I guess that I should differentiate $e^{\tan (t)}$ first, the derivative of $f(x)=e^x$ is $f'(x)=e^x$, but I got confused at how to proceed later, I've differentiated part of the $\tan x$ in a similar fashion to this. At the moment, my guess is that I should finishing doing the derivative of $\tan (x)$, which is $\sec ^2(x)$ then apply the chain rule, obtaining:

$$e^{\tan (t)}\sec^2(t)$$

Now I guess I should use the product's rule to multiply $t$ by $e^{\tan (t)}\sec^2(t)$, obtaining:

$$1 \cdot e^{\tan (t)}+t\cdot e^{\tan (t)}\sec^2(t)$$

And now I guess I should use the quotient rule on:

$$\frac{1 \cdot e^{\tan (t)}+t\cdot e^{\tan (t)}\sec^2(t)}{ln(3t+1)}$$

But I feel this is getting too lengthy and that perhaps I'm doing something wrong. Is my reasoning until this point correct?

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  • $\begingroup$ What do you mean "derive?" Maybe differentiate? $\endgroup$ – ClassicStyle Jul 10 '14 at 23:17
  • $\begingroup$ @TylerHG Yes. It was a poor adaptation from portuguese. I thought one could employ "derive" in english. $\endgroup$ – Billy Rubina Jul 10 '14 at 23:18
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    $\begingroup$ It's ok. Derive means to take something and get something new from it. Like derive a closed form expression for the integral of this function $\endgroup$ – ClassicStyle Jul 10 '14 at 23:22
  • $\begingroup$ @TylerHG Thanks for the heads up. $\endgroup$ – Billy Rubina Jul 10 '14 at 23:23
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The method you are using is correct up until you try to use the quotient rule. The mistake is that you already took the derivative of the numerator. The quotient rule is:

$$\frac{d}{dx}\frac{f}{g}=\frac{f'g-fg'}{g^2} $$

Some times it's easier to cast the expression as a product instead of a quotient...i.e.

$$ \frac{d}{dt}te^{\tan(t)}(\ln(3t+1))^{-1} $$

And then just use the product rule.

EDIT

I will first demonstrate the method for using the quotient rule.

$$ \frac{d}{dt}\frac{te^{\tan(t)}}{\ln(3t+1)}=\frac{\left(te^{\tan t}\right)'\ln(3t+1)-te^{\tan t}\left(\ln(3t+1) \right)'}{\ln^2(3t+1)} $$

Then, using the chain rule and product rule on the first and second derivative terms in the numerator (chain and product rule within the quotient rule):

$$ \frac{\left(e^{\tan t}+te^{\tan t}\sec^2t\right)\ln(3t+1)-\frac{3te^{\tan t}}{3t+1}}{\ln^2(3t+1)} $$

We can then go on to simplify the expression which I leave you to do.

An alternate method is casting the entire expression as a product like I did in the original post. Here we would have using the chain and product rule:

$$ \frac{d}{dt}te^{\tan t}(\ln(3t+1))^{-1}=e^{\tan t}(\ln(3t+1))^{-1}+te^{\tan t}\sec^2(t)(\ln(3t+1))^{-1}\dots-te^{\tan t}(\ln(3t+1))^{-2}\frac{3}{3t+1} $$

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  • $\begingroup$ So I should use one of them or the other but not both? $\endgroup$ – Billy Rubina Jul 11 '14 at 14:32
  • $\begingroup$ What do you mean $\endgroup$ – ClassicStyle Jul 11 '14 at 15:16
  • $\begingroup$ I used the chain rule, using that approach, what should be done next? $\endgroup$ – Billy Rubina Jul 11 '14 at 17:26
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    $\begingroup$ The problem is that you took the derivative of the numerator using the chain and product rules and then formed a new fraction with the derivative of the old numerator in the new numerator. This is wrong. You need to either use the chain and product rules within the quotient rule using the formula for the quotient rule or use just one large product rule. $\endgroup$ – ClassicStyle Jul 11 '14 at 18:43
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    $\begingroup$ I will make an edit showing both methods. Give me a second $\endgroup$ – ClassicStyle Jul 11 '14 at 18:56
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You can make this problem easier by logarithmic differentiation. If

$$y=\frac{t e^{\tan(t)}}{\ln(3t+1)}$$

then using the rules of logarithms

$$\ln(y) = \ln(t) + \tan(t) - \ln(\ln(3t+1)).$$

This is easier to differentiate than what you started with, because you don't have to do anything with the product or quotient rules. When you differentiate this, you get $\frac{y'}{y}$, so

$$y' = y \frac{d}{dt} \left ( \ln(y) \right )$$

A caveat: this only works for $t>0$, whereas the given function is also defined and differentiable when $-1/3 < t < 0$. You can still use this method for $t<0$ by writing $\frac{t}{\ln(3t+1)} = \frac{-t}{-\ln(3t+1)}$ and then doing the same procedure.

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  • $\begingroup$ Thanks for the help. But the way I was doing would also work, wouldn't it? (No, I'm not planning to use it, I'm just making the stupidity test on myself). $\endgroup$ – Billy Rubina Jul 10 '14 at 23:22
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    $\begingroup$ If you were going to do it that way, I would organize it the other way. That is, the whole problem is a quotient rule problem with $f(t)=t e^{\tan(t)}$ and $g(t)=\ln(3t+1)$, so the final result is $\frac{f'g - g'f}{g^2}$. Differentiating $f$ is a product rule problem. Differentiating $g$ and $e^{\tan(t)}$ are chain rule problems. $\endgroup$ – Ian Jul 10 '14 at 23:28
  • $\begingroup$ Then I didn't need to use the chain rule on $e^{\tan(x)}$? $\endgroup$ – Billy Rubina Jul 10 '14 at 23:32
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    $\begingroup$ No, you did. When you go to compute $f'$, you need to compute $d/dt(t)$ and $d/dt(e^{\tan(t)})$, so that you can use the product rule. The latter needs the chain rule. $\endgroup$ – Ian Jul 10 '14 at 23:33
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$$\begin{equation}\begin{split}f(t)=\dfrac{t\exp(\tan t)}{\log(3t+1)}=\dfrac{g(t)}{h(t)},\end{split}\end{equation}$$

$$\begin{equation}\begin{split}\dfrac{\mathrm{d}f(t)}{\mathrm{d}t}&=\dfrac{\dfrac{\mathrm{d}g(t)}{\mathrm{d}t}h(t)-\dfrac{\mathrm{d}h(t)}{\mathrm{d}t}g(t)}{h^2(t)},\\&=\dfrac{\mathrm{d}g(t)}{\mathrm{d}t}\dfrac{1}{h(t)}-\dfrac{\mathrm{d}h(t)}{\mathrm{d}t}\dfrac{g(t)}{h^2(t)}.\end{split}\end{equation}$$

Now: $$\dfrac{\mathrm{d}g(t)}{\mathrm{d}t}=\exp(\tan t)+t(1+\tan^2t)\exp(\tan t).$$ $$\dfrac{\mathrm{d}h(t)}{\mathrm{d}t}=\dfrac{3}{3t+1}.$$

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While others have already mentioned a couple of different approaches that can be applied to your problem, I wanted to emphasize how top-down thinking can be applied to solve problems like this.

We start with the quotient $\dfrac{te^{\tan t}}{\ln(3t+1)}$; this is of the form $\frac fg$, where $f(t) = te^{\tan t}$ and $g(t) = \ln(3t+1)$. We know that the result will be $\frac{gf'-fg'}{g^2}$, so we know that we need to compute $f'$ and $g'$.

Let's look at $f$ first: $f(t) = te^{\tan t}$. This has the form $f=f_1f_2$, where $f_1(t) = t$ and $f_2=e^{\tan t}$, and the derivative will be $f'=f_1'f_2+f_2'f_1$, so we need to compute $f_1'$ and $f_2'$. Fortunately, $f_1'$ is trivial: $\frac{df_1(t)}{dt} = 1$.

Meanwhile, $f_2$ has the form $f_2=f_{21}(f_{22}(t))$, with $f_{21}(t) = e^t$ and $f_{22}(t) =\tan t$; the derivative will be $f_2'(t) = f_{22}'(t)f_{21}'(f_{22}(t))$, so we need to find the derivatives of $f_{21}$ and $f_{22}$.

And so on: you can turn this into a completely mechanical procedure for finding the derivative of an (elementary) function by breaking it repeatedly into constituent pieces. Once you've found the derivatives of all of the individual chunks, you can start reassembling them to find the derivatives of your intermediate pieces, etc.; it can even help to write this out physically as a tree so that you don't lose track of what to do with any given piece.

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  • $\begingroup$ Excuse me, you wrote "$f_2'\cdot f_1'$ is trivial", shouldn't it be $f_2\cdot f_1'$? $\endgroup$ – Billy Rubina Jul 12 '14 at 22:59
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    $\begingroup$ @IgäriaMnagarka That's actually supposed to be an end-of-sentence, not a product. :-) I'll juggle things to make it much clearer. $\endgroup$ – Steven Stadnicki Jul 12 '14 at 23:00

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