Question 1

i am confused to what explicit and implicit mean

Accepted Answer

With explicit differentiation, you're deriving a new function from an existing function. That is, given `f(x)`, you're generating `f'(x)`. That has a big limitation: it has to be a function (something as simple as a circle won't work) and there can only be one variable.

With implicit differentiation, you're transforming expressions. `d/dx` becomes an algebraic operation like sin or square root, and can perform it on both sides of an equation.

Implicit differentiation is a little more cumbersome to use, but it can handle any number of variables and even works with inequalities. Generally, if you can learn implicit differentiation, you can forget explicit because you can always just do `dy/dx = df(x)/dx` to find `f'(x)`.

Question 2

I understand the mechanics here, I believe I get that this works but not why. The part I am confused about is we now suddenly seem to be doing math WITH operators. I though the d/dx was just an operator symbol, but now we seem to be treating like a number or variable and can multiply or divide with it?!

Clearly I am misunderstanding, but to me it's like suddenly learning " (+) * (+) = +^2", addition sign times addition sign of course equals addition sign squared".

Clearly I am misunderstanding, but to me it's like suddenly learning " (+) * (+) = +^2", addition sign times addition sign of course equals addition sign squared".

Accepted Answer

There is no spoon!

Yes, d/dx is an operator, but not like + 
Other than the fact that + is diadic and d/dx is monadic (which has nothing to do with the discussion at hand, but it is a difference between the two) it is best to think of d/dx as an operator that takes a function as input, and returns a function. In this light it may be easier to see how it is being manipulated. This will be even more clear in Integral calculus when you do u-sub problems. Later you will be seeing the same thing with other operators you will be learning in multivariable calculus.

Do your best to accept it and use it (or do your own research on what d/dx, in all its subtle glory, is). When I was a student I used to say that the stuff I was taught in one semester would be understandable (from a theorem point of view, not a use point of view) in the following semester, by which time, our maturity grows so the we can better accept the truth. (which is to say, sometimes we need more time to understand the math within the proofs of the tools we are using, but that we don't understand at this entry level of our math studies, doesn't mean the concepts don't work; a rigorous proof of why does exist)

Question 3

It seems to me like it's more work to use implicit differentiation than it is explicit differention. Is there any good reason to use implicit differentiation?

Accepted Answer

Sure, you cannot always isolate y or else doing so is nightmarishly difficult. In these cases implicit differentiation is much easier.
 
For example, try finding the derivative of this by explicit differentiation:
y=ln(y+x)

Question 4

at 0:59, is dy/dx the same thing as d/dx [x^-2] because y = x^-2?

Accepted Answer

yes that's how you write the notation. You can also write dy/dx as (y)' or f'(x). There are many ways in which you can indicate a derivative of a function.

Question 5

So in the end, to solve for y, we still had to use the result from explicit differentiation? So is it impossible to solve for y using only implicit differentiation?

Accepted Answer

Sometimes you cant solve explicitly. The original equation however could be, he was just using it as an example to prove the explicit and implicit both provide the same answer. However, when you solve implicitly, you get a y variable in the answer, which would have to be replaced with the y= from the original equation. When you do that and simplify, both answers are the same.

Question 6

Is there a way to explain implicit differentiation in a way other than “it is an application of the chain rule”? Can it be explained in terms of limits? I am not feeling comfortable with implicit differentiation. I can apply the rules and get the correct answers, but I don’t have a gut feeling for why it works. It is hard for me to explain. Please bear with me.

It feels like a leap to me to say that the chain rule fully explains implicit differentiation. Everything I’ve learned so far about differentiation has been based on explicitly defined functions and limits. Applying the chain rule to explicit functions makes sense to me, as I am just recognizing composite functions within an original function. But applying the chain rule to a non-function, e.g. an equation of a circle, and to the dependent variable, seems like a giant leap. Take for example, the equation x√y=1.  I understand that y is a function of x, but it is the given function that makes it a function of x. Solving the equation for y yields y=1/x^2 . If I substitute 1/x^2  in for y in the original equation, I get 1=1. This is different than the equation y=xsin(2x^2+2x+1) where 2x^2+2x+1 is a composite function.

Perhaps I am not completely understanding the chain rule. What am I missing? Thanks for any help you can provide.

Accepted Answer

Thinking of d/dx as an operator that accepts functions and returns their derivative is perfectly valid, and you derivation of d/dx[x]=1 is valid.

Question 7

I suppose it depends on the equation...but can you just look at an equation and know that it can't be derived using explicited differentiation? Is there a sort of 'telltale' that lets you know that "okay, this literally can't be solved explicitedly, and must be implicitedly derived?"

Accepted Answer

If you can't solve the equation for y, you need implicit differentiation.

Question 8

Im confused at the end of the video when Sal explains how explicit and implicit differentiation are related. Could someone clarify the relationships and similarities of the two?

Accepted Answer

Finding the derivative explicitly is a two-step process: (1) find y in terms of x, and (2) differentiate, which gives us dy/dx in terms of x.

Finding the derivative implicitly is also two steps: (1) differentiate, and (2) solve for dy/dx. This method may leave us with dy/dx in terms of both x and y. However, we previously found y in terms of x when we found the derivative explicitly. We have an expression in x that is equal to y, so we can substitute that expression for y anywhere y appears in the derivative we found implicitly. The result will be dy/dx in terms of x, eliminating any y that appears on the right side of the equation. If we did everything right, this derivative will be the same as the one we found explicitly.

Question 9

What supports Sal squaring both sides at :30

Accepted Answer

That's basic algebra. If a = b, then a^2 = b^2. You're multiplying both sides of the equation by amounts that are known to be equal (a and b in my example, or sqrt(y) and 1/x in the video), so the equality is preserved.

Question 10

@ 2:42 Sal uses the chain rule.  Do you always have to use the chain rule in implicit differentiation?

Accepted Answer

Yes, implicit differentiation is a special application of the chain rule. It's how we take the derivative of an expression involving y with respect to x, which otherwise doesn't sound possible (we normally need a function of x in order to differentiate with respect to x).

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 3

Showing explicit and implicit differentiation give same result

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Video transcript