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## Differential Calculus

### Unit 3: Lesson 4

Sal finds dy/dx for e^(xy²)=x-y using implicit differentiation. Created by Sal Khan.

## Want to join the conversation?

• For those who get stuck at . I spent a few hours trying to figure out how Sal got e^xy^2 in the form he did to get the derivative. I finally found a sheet on derivatives of powers of e with the following: the derivative of powers of e is equal to e and it's exponent and/or power (times) the derivative of the exponent and/or power. I had not see this explained in any of Sal's work prior to this video so I was quite stuck (hey, this is the absolute 1st time I've done calculus so I had no clue). I hope this helps anyone who has the same question
--I thought it would have been nice for Sal to include this info regarding derivates of powers of e, but I guess he has a somewhat difficult task of deciding just how basic to explain certain things. Since I had not seen this explained prior to this point in the calculus videos (and I've watched them all up to here)......I thought this would be nice info to include here for anyone needing it --thx
• This is an application of the chain rule where you take the derivative of the outer equation which is e^x (where x=the inner function) and times that by the derivative of the inner function which is xy^2 where you'll have to use the product rule to get that derivative. Also you have to think of y as a function of x, that is why he uses the chain rule again on y^2, where the outer function is (something)^2 and then you take the derivative of the inner function (y) with respect to x getting dy/dx or in case of the notation used in this video, y'
• At he discusses notation and why with the capital D it is not clear whether if you are taking the derivative with respect to X. But one thing in all these videos that has not been clear to me: why with respect to X? What happens if you go the other way? Does it still work (but with a lot more hairy math)? I'm sure there is something basic I am not seeing, but any thoughts would be great.
Cheers...
• I think the reason that we usually differentiate with respect to x is because this gives us the slope. And this is only because by convention x is normally the independent variable and shown on the horizontal axis when graphed.
Remember that differentiation is about the rate of change of a function with respect to some variable. dy/dx means the change in y with respect to the change in x. dy/dx = rise/run = slope. If we differentiated with respect to y (dx/dy) then we would know the change in x for a given change in y, which would be the run/rise, or reciprocal of the slope. Not any more complicated; just not as useful.
It might help to consider that in a lot of physics problems, the derivative is with respect to t (time), rather than x, because we are interested in how something changes with time. If, say, we were looking for how a horizontal position 'x' changed with time 't' then we would be interested in dx/dt.
• at , when we add y' to each side, we do not have 2y'. The only difference is that the y' on the left just disappears. Why is this?
• Think of it like this:
blah*y' + 1*y' = (blah + 1)y' this is the same thing, distributive property
and Sal added +1y' to both sides of the equation to get rid of the y' on the right side of the equation and you can do that because doing the same thing to both sides of the equation doesn't change the equality of the equation.
• in the final solution, can you plug in x-y for the e^(xy^2) since the original equation says that they are equal? this would give a slightly less hairy expression i think.
• Yes, I think you can do this. Sal does a similar thing in the next video (Derivative of x^(x^x)). Can someone more knowledgeable than I please confirm this?
• at why does he get "+1)y'" and not just 2y'? it seems like there is already a y' on that side so now that he added another there should be two right?
• He just skipped one step, when he added y' he got
2xy(e^xy^2)*y'+y'
Now you can't just add this and get 2y' but they both have a y' so you can do this
y'(2xy*e^xy^2+1)
• At you multiply by Y' but why is the reason for this? If the product rule is just f'(x)g(x)+f(x)g'(x) which would mean from my understanding (1)(y^2) + (x)(2y)
• he multiplies by y prime because each time you take the derivative of y you have to add a y prime.
(1 vote)
• Can someone PLEASE help me here. I have this problems from the Implicit Differentiation practice section, which is the exercise right after this one, but I am EXTREMELY CONFUSED.
OK, so here is the problem:
Find dy/dx for cos^2(xy)=x+y,
Ok, So the next step would be:
2cos(xy)*(-sin(xy))*(y+x(dy/dx))=1+dy/dx.
Ok cool, I got to this part.
BUT THEN the solution does this, WHICH I HAVE NO IDEA WHYYYYYYY!
2ycos(xy)*(-sin(xy))+2xcos(xy)*(-sin(xy))dy/dx=1+dy/dx.
Someone NEEDS to explain this ^ part to me and how it became to be this or else I will BLOW MY BRAINS OUT!
• The derivative of y is dy, of x is dx. So that was main mistake. Note: It is easier to do these problems with the y' notation instead of the dx notation.
Here is how to solve the problem:
``cos²(xy)=x+y2cos(xy) (-sin(xy)) (xdy + y dx) = dx + dy−2cos(xy) sin(xy) (xdy + y dx) = dx + dy−2x cos(xy) sin(xy) dy −2y cos(xy) sin(xy) dx = dx + dy−2x cos(xy)(sin(xy) dy − dy = dx + 2y cos(xy) sin(xy) dx dy [−2x cos(xy) sin(xy) − 1 ] = dx [ 1 + 2y cos(xy) sin(xy) ]dy (−1)[ 1+2x cos(xy) sin(xy)] = dx [1 + 2y cos(xy) sin(xy)]dy = (−1)[1 + 2y cos(xy )sin(xy)] dx / [1+2x cos(xy) sin(xy)] dy/dx =  (−1)[1 + 2y cos(xy) sin(xy)] / [1+2x cos(xy) sin(xy)] ``

You can leave it like that or you can use the following property to simplify:
2 sin(u) cos(u) = sin(2u)
Thus,
``dy/dx = (-1) { y sin(2xy) + 1 ] / [ x sin(2xy) + 1]``
(1 vote)
• Would it be ok to rewrite the original relationship as:
xy^2 = ln(x -y)
I think this is slightly less messy to differentiate.
• It will be a valid operation but you'll end up with a different derivative for a different relation. See for yourself - http://www.wolframalpha.com/input/?i=x*y%5E2+%3D+ln(x-y)
• Could I take natural logarithm on the both sides and then do differentiation on xy^2 = ln(x-y) instead of differentiating e^xy^2 = x-y directly?
• In line 3, instead of multiplying e^xy^2 to each term in the brackets, i took it to the right side and i took the derivative from the right side to the left. So the equation became...
y^2+(2xy)(y')+y'=1/e^xy^2
(y')(2xy)+y'=(1-(y^2)(e^xy^2))/e^xy^2
y'(2xy+1)=(1-(y^2)(e^xy^2))/e^xy^2
therfore
y'=(1-(y^2)(e^xy^2))/(2xy)(e^xy^2)+e^xy^2

where did i go wrong?