How can you explain implicit differentiation?

Question

So I am taking calculus 1 online from a local college (bad idea, but the only thing that fit my schedule).

The professor used the notation $f'(x) =$ for EVERY function up until two weeks ago. All of the sudden he changed his notation to d/d$x$, d$y$/d$x$ and hasn't been able to explain it clearly enough via videos.

I am trying hard to learn how implicit differentiation works, and I've been putting in six hours per day of studying for two weeks. I just don't get it. I have read and tried the examples on MIT opencourseware, our textbook and calculus for dummies.

What is this d/d$x$ that gets placed everywhere? Where do you place it? Why use it? Same with $\text{d}y/\text{d}x$.

I've had lots of success here, so if someone could show me how to do this i would really appreciate it. This is my last hope :)

Thanks

Answer 1

To answer your first question,

$\frac d{dx}$ is an operator. So let your function be $f(x)$. The notation $\frac d{dx}f(x)$ is equivalent to saying "the derivative of $f$ with respect to $x$". Often times we will see $\frac {dy}{dx}$. This is similar. It is just "the derivative of a function of the $y$ value with respect to $x$".

For the next question,

implicit differentiation is used when it is not necessarily easier to simplify the equation first. However, we can differentiate the equation implicitly and then solve for the derivative. For an easy example:

$y^2 + x = 1$ apply $\frac d{dx}$

$\frac d{dx} (y^2 +x) = \frac d{dx} (1)$

Here the function $y$ is actually a function dependent on $x$. So we would formally write $y(x)$.

So, when we take the derivative we can simply use product rule!

$\frac d{dx}(y(x) \cdot y(x)) = y'(x)\cdot y(x) + y(x)\cdot y'(x) = 2y(x)y'(x)$

And then differentiate all other terms with respect to $x$.

$2yy' + 1 = 0$

Now it is possible to solve for $y'(x)$ which was what we were looking for! This is also the "derivative of $y(x)$ with respect to $x$" which we just learned is equivalent to $\frac d{dx} {y(x)}= \frac {dy}{dx}$!

Hopefully this cleared some things up.

Answer 2

What is the derivative of a function? It's measure of the slope, or rate of change of a function. So what is $f'(x)$? it's the limit of the ratio of difference in height (difference in $f(x)$), to the difference in $x$, as the differences go to zero.

If you think of the $d$ in $dx$ as shorthand for "difference", then you immediately see that $df/dx$ is a very suggestive notation for $f'(x)$.

$\quad\quad\quad\quad\quad\quad\quad$

Now, what about implicit differentiation? Even though $df/dx$ is not a real fraction, just a good notation, pertending it is a fraction leads to the law of implicit differentiation: $$\frac{df}{dx}=\frac{df}{dy}\frac{dy}{dx}$$ Or in our older notation: $$f'\left(y(x)\right) \text{(derivative acc. to $x$)}=f'(y) y'(x)$$ Now you see the strength of the $df/dx$ notation - there is no confusion about what variable the function is being derived by.