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# Showing explicit and implicit differentiation give same result

Sal gives an example of how implicit differentiation results in the same derivative expression as direct differentiation. Created by Sal Khan.

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• 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? • 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". • 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)
• at , is dy/dx the same thing as d/dx [x^-2] because y = x^-2? • 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? • 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. • • • 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? • 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.  