How to differentiate $x^4y^4$ with respect to $x$, if $y$ is a function of $x$?

Question

I have an online homework question that reads: differentiate the expression $x^4y^4$ with respect to $x$. The matching homework question in the textbook reads the same except the exponents are different: $x^2y^3$.

For the textbook question I followed the product rule and got $(2xy^3)(3y^2)$. However, the answer key says the answer is $\frac{d}{dx}(x^2y^3)=3x^2y^2y'+2xy^3$. Can someone explain the methods to achieving this method and why it is correct. Also what exactly does $\frac{d}{dx}$ mean?

Answer 1

Implicit differentiation follows from the chain rule.

You know that if we differentiate $\sqrt{x^{3}}$, this is really $(x^{3})^{\frac{1}{2}}$, so we have to use the chain rule to get:

$\dfrac{d}{dx}((x^{3})^{\frac{1}{2}}) = [\frac{1}{2}(x^{3})^{-\frac{1}{2}}] \cdot 3x^{2}$

Well, what if I wrote $y = x^{3}$ and just told you to differentiate $\dfrac{d}{dx} (y)^{\frac{1}{2}}$? Note that I am asking you to differentiate the function $y$ with respect to the variable $x$, but in the formula $(y)^{\frac{1}{2}}$, you don't see the variable $x$. So you tell yourself "I know $y$ is a function of $x$ even though I don't see the variable $x$, so I will apply the chain rule as I would if I did see the $x$":

We get: $\dfrac{d}{dx} (y)^{\frac{1}{2}} = \frac{1}{2} (y)^{-\frac{1}{2}} \cdot \frac{dy}{dx}$ where $dy/dx$ is the derivative of the function $y$ with respect to $x$ which me multiply as a result of the chain rule (remember, we are thinking here that $y$ represents a function of $x$ -- and we are doing this implicitly because nowhere do you see the variable $x$ in the name $y$ -- in fact, I wrote $y = x^{3}$ earlier to tell you exactly what the function of $x$ is, but we didn't use it to differentiate implicitly).

Now, to check that the implicit differentiation did the job correctly, since I know exactly what $y$ is (it is $x^{3}$), we get

$\dfrac{d}{dx} (y)^{\frac{1}{2}} $

$= \frac{1}{2} (y)^{-\frac{1}{2}} \cdot \frac{dy}{dx}$

$= \frac{1}{2} (x^{3})^{-\frac{1}{2}} \cdot 3x^{2}$

And this agrees with the differentiation we did earlier where it wasn't implicit!

Answer 2

You should treat $y=f(x)$ for some function $f$. Therefore, the expression $x^2y^3=x^2[f(x)]^3$. When you differentiate this expression, you use the product rule and the chain rule. For example, \begin{align*} \frac{d}{dx}(x^2y^3)&=\left(\frac{d}{dx}x^2\right)y^3+x^2\left(\frac{d}{dx}y^3\right)\\ &=2xy^3+x^2\left(3y^2\frac{dy}{dx}\right)\\ &=2xy^3+3x^2y^2y' \end{align*}

This is equivalent to the calculation \begin{align*} \frac{d}{dx}\left(x^2[f(x)]^3\right)&=2x[f(x)]^3+x^2(3[f(x)]^2f'(x))\\ &=2x[f(x)]^3+3x^2[f(x)]^2f'(x). \end{align*}