# Math Help - Unsure of how to differentiate this expression

1. ## Unsure of how to differentiate this expression

Hi everyone,

this is an exercise from my circuit class, it involves calculus, and I am unsure about how to do it.

Differentiate the following expression with respect to RL:

From the solutions manual it seems like I am supposed to use the quotient rule. This confuses me, because (other than being a noob in math) I know that one of the variables is R, as specified in the exercise, I imagine RL as being "x" in a "normal" differentiation problem. Other than the RL being kind of an "x" I only see constants in this expression, so why would I use the quotient rule? Isn't the quotient rule only for expressions which have two functions divided by each other? Does this mean that VTH is some kind of function?

I will give a picture of the solution too, so that you can take a look at it thank you all for reading!

2. ## Re: Unsure of how to differentiate this expression

The expression is like this

$\frac{V_{TH}^2R_L}{(R_{TH}+R_L)^2}$

Which if you rewrite in the variable x with constants a, b

$\frac{a^2x}{(b+x)^2}$

So you can see it makes sense to use the quotient rule

3. ## Re: Unsure of how to differentiate this expression

Nora, You have to use product rule, because you have the product (multiplication) of two functions u and v of $R_L$one is $u=(\frac{V_{TH}}{R_{TH}+R_L})^2$ and the other is just $v=R_L$. Let me refresh product rule, $(uv)' = u'v +v'u$ it's clear that $v' =1$. Now, to get u' we should use chain rule, since we have a function inside of another, let me refresh that too $(f(g(x))' = f '(g(x)) * g'(x)$ So you basically derive the outside function (that is $x^2$ where x is the fraction without the exponent) and you get 2x for that, then you multiply by the derivative of the inside function, so we need the derivative of $\frac{V_{TH}}{R_{TH}+R_L}$ for this one, you can use quotient rule $(\frac{f}{g})' = \frac{f'g-g'f}{g^2}$ where $f(R_L)=V_{TH}$ and $g(R_L)=R_{TH}+R_L$, for this, $f'(R_L) = 0$ and $g'(R_L)=1$ so this is just $\frac{-V_{TH}}{(R_{TH}+R_L)^2}$. Putting all that together, $u' =2\frac{V_{TH}}{R_{TH}+R_L} *\frac{-V_{TH}}{(R_{TH}+R_L)^2}$ we can simplify all this to just $u' = \frac{-2V_{TH}^2}{(R_{TH}+R_L)^3}$. Now that we have all we need, we put them together as product rule says, so we would have : $P'(R_L) = \frac{-2V_{TH}^2}{(R_{TH}+R_L)^3}*R_L + 1 * (\frac{V_{TH}}{R_{TH}+R_L})^2$ If we do a common denominator and factor out the $V_{TH}^2$ we get $P'(R_L) = (V_{TH})^2\frac{(-2R_L+ R_{TH}+R_L)}{(R_{TH}+R_L)^3}$ that is just $P'(R_L) = (V_{TH})^2\frac{(R_{TH}-R_L)}{(R_{TH}+R_L)^3}$ The answer they gave you is not simplified, it is equivalent to this one though, just factor out $(R_L+R_{TH})$ in the numerator and cancel it with one of the denominator's. I hope that helped, regards
Damián Vallejo.
You can visit my blog if you want, it is Math 911 I expose various math topics there.

4. ## Re: Unsure of how to differentiate this expression

Note that Shakarri is saying "you have to use the quotient rule" and chalaman1403 is saying "you have nto use product rule"! You can, in fact, use both because you can think of the quotient $\frac{f(x)}{g(x)}$ as being the product $f(x)g^{-1}(x)$. Whether it is simpler to use the quotient rule (which is a little more complicated than the product rule alone) or use the product rule and a chain rule to handle the negative one power, is mainly a matter of taste.

5. ## Re: Unsure of how to differentiate this expression

Thank you so much everyone! When you guys put it so nicely it seemed way easier than before. I got it now.