Thread: Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

1. Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

Hello,

I just graduated and now feel masochistically inclined to torture myself with diff eq. My question relates to differentiating terms in implicit functions using the product rule. I understand how to use the product rule with two differentiable functions of x. I do not, however, quite understand how to use this rule when the terms involve y and x.

I have attached a screenshot of the problem. I have highlighted in red the term whose origin I do not quite understand.

f(x) = y^4 + 2x²y² + 6x² = 7

u = 2x²y²
v = y²

The derivative of 2x²y² w.r.t x is 2y(dy/dx). But why?

Question: Why does dy/dx appear from u(dv/dx)? Why is the derivative not simply 2y?

Thank you for any help,
postg

2. Re: Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

In my opinion what is not full explained is the fact that y isn't an independent variable, but is a function of x, so that if $\displaystyle u(x)=2\ x^{2}$ and $\displaystyle v(x)= y^{2}(x)$ is...

$\displaystyle u\ \frac{d v}{d x} = 2\ x^{2}\ 2\ y(x)\ \frac{dy}{dx}= 4\ x^{2}\ y\ y^{'}$ (1)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. Re: Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

Why do you define v(x) = y²(x)?
The problem defines v = y², not v = y²(x). What am I missing?

I'm afraid I don't understand your answer to the question. Why is the derivative of y² = 2y(dy/dx). Why not just 2y, by the nx^n-1 rule?

4. Re: Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

Originally Posted by postg
Why do you define v(x) = y²(x)?
The problem defines v = y², not v = y²(x). What am I missing?

I'm afraid I don't understand your answer to the question. Why is the derivative of y² = 2y(dy/dx). Why not just 2y, by the nx^n-1 rule?
There is also the CHAIN RULE.
The derivative of (x^2), with respect to x, is 2x(dx/dx). dx/dx is ONE.
The derivative of (y^2), with respect to x, is 2y(dy/dx). dy/dx is NOT ONE!

5. Re: Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

Originally Posted by postg
Why do you define v(x) = y²(x)?
The problem defines v = y², not v = y²(x). What am I missing?

I'm afraid I don't understand your answer to the question. Why is the derivative of y² = 2y(dy/dx). Why not just 2y, by the nx^n-1 rule?
It's because $\displaystyle \displaystyle y$ is a function of $\displaystyle \displaystyle x$. The derivative is taken with respect to $\displaystyle \displaystyle x$, not $\displaystyle \displaystyle y$.

You want to find the derivative of $\displaystyle \displaystyle v = [y(x)]^2$, in other words $\displaystyle \displaystyle \frac{dv}{dx}$.

Let $\displaystyle \displaystyle w = y(x)$ so that $\displaystyle \displaystyle v = w^2$. From the chain rule, we would know that $\displaystyle \displaystyle \frac{dv}{dx} = \frac{dw}{dx} \cdot \frac{dv}{dw}$.

$\displaystyle \displaystyle \frac{dw}{dx} = \frac{d}{dx}[y(x)] = \frac{dy}{dx}$.

$\displaystyle \displaystyle \frac{dv}{dw} = 2w = 2y$.

So $\displaystyle \displaystyle \frac{dv}{dx} = 2y\frac{dy}{dx}$.

6. Re: Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

Originally Posted by postg
Why is the derivative of y² = 2y(dy/dx). Why not just 2y, by the nx^n-1 rule?
Look at the word implicit. When asking for the derivative of $\displaystyle y^2$ with respect to $\displaystyle x$, we are implying (it is implicit) that $\displaystyle y^2$ is a function $\displaystyle x$,

Thus $\displaystyle \frac{d(xy^2)}{dx}=y^2+2xy\frac{dy}{dx}.$