Implicit differentiation word problem

I have a written assignment in which one specific word problem is confusing me greatly. I just recently learned implicit differentiation but haven't gotten much into depth with it. The question is as follows:

"Each time we use implicit differentiation, we are able to rewrite the resulting equation in the form:

f(x,y)=y′ ·g(x,y)

for some expressions f and g.

Explain why this can always be done; that is, why doesn’t the chain rule ever produce a term like:

(y′)^2

or

1/y′ "

I really am not grasping the form of the initial equation given let alone the logic behind what it is asking. I don't want anyone to give me the answer to the problem, but any help in understanding what it is actually asking, the concepts behind the question and the thought process required to answer the question would be of great help.

Re: Implicit differentiation word problem

$\displaystyle F(x,y(x))=0\Rightarrow \frac {\partial F}{\partial x}+\frac {\partial F}{\partial y}y'(x)=0\Rightarrow \ldots $

Re: Implicit differentiation word problem

Quote:

Originally Posted by

**elrpsu** I have a written assignment in which one specific word problem is confusing me greatly. I just recently learned implicit differentiation but haven't gotten much into depth with it. The question is as follows:

"Each time we use implicit differentiation, we are able to rewrite the resulting equation in the form:

f(x,y)=y′ ·g(x,y)

for some expressions f and g.

Explain why this can always be done; that is, why doesn’t the chain rule ever produce a term like:

(y′)^2

or

1/y′ "

I really am not grasping the form of the initial equation given let alone the logic behind what it is asking. I don't want anyone to give me the answer to the problem, but any help in understanding what it is actually asking, the concepts behind the question and the thought process required to answer the question would be of great help.

Every function of y is a function of x. So if you wanted to differentiate something like $\displaystyle \displaystyle z= y^2$ (in other words, $\displaystyle \displaystyle z = \left[y(x)\right]^2$ )with respect to x, you need to use the chain rule.

The "inner" function is $\displaystyle \displaystyle y(x)$, and the "outer" function is $\displaystyle \displaystyle y^2$.

So using the chain rule, if $\displaystyle \displaystyle z = y^2$ and we let $\displaystyle \displaystyle u = y \implies z = u^2$, then

$\displaystyle \displaystyle \frac{du}{dx} = \frac{dy}{dx} $ and $\displaystyle \displaystyle \frac{dz}{du} = 2u = 2y$.

Therefore $\displaystyle \displaystyle \frac{dz}{dx}= \frac{dz}{du} \cdot \frac{du}{dx} = 2y\,\frac{dy}{dx}$.

So whenever you want to differentiate a function of y, since y is the inner function, you are always going to get $\displaystyle \displaystyle \frac{dy}{dx}$ as a factor because of the chain rule.

Re: Implicit differentiation word problem

Quote:

Originally Posted by

**Prove It** Every function of y is a function of x. So if you wanted to differentiate something like $\displaystyle \displaystyle z= y^2$ (in other words, $\displaystyle \displaystyle z = \left[y(x)\right]^2$ )with respect to x, you need to use the chain rule.

The "inner" function is $\displaystyle \displaystyle y(x)$, and the "outer" function is $\displaystyle \displaystyle y^2$.

So using the chain rule, if $\displaystyle \displaystyle z = y^2$ and we let $\displaystyle \displaystyle u = y \implies z = u^2$, then

$\displaystyle \displaystyle \frac{du}{dx} = \frac{dy}{dx} $ and $\displaystyle \displaystyle \frac{dz}{du} = 2u = 2y$.

Therefore $\displaystyle \displaystyle \frac{dz}{dx}= \frac{dz}{du} \cdot \frac{du}{dx} = 2y\,\frac{dy}{dx}$.

So whenever you want to differentiate a function of y, since y is the inner function, you are always going to get $\displaystyle \displaystyle \frac{dy}{dx}$ as a factor because of the chain rule.

Thank you! That helped me understand this all quite a bit better than I was able to before.

Re: Implicit differentiation word problem

Another way of expressing it is that differentiation is a **linear** operation: (f+ g)'= f'+ g' and (fg)'= f'g+ fg'. You never get "squares" or more complicated functions of the derivative.