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## Differential Calculus

### Course: Differential Calculus>Unit 3

Lesson 4: Implicit differentiation (advanced examples)

Implicit differentiation of y = cos(5x - 3y). Created by Sal Khan.

## Want to join the conversation?

• In this video at as well as in the last video, sal subtracts something from both sides that contains dy/dx, in order to get all dy/dx to the same side. But I don't get how one of the dy/dx becomes 1. In this video it results in (1-3sin(5x-3y))dy/dx. How does dy/dx-3sin(5x-3y)*dy/dx= (1-3sin(5x-3y))*dy/dx •  He factored the dy/dx out. For example, in the expression x+5x we can factor out the x to get (1+5)x. The 1 is left behind to show that we had an x there before we factored. Another way to see this is to try multiplying the dy/dx back into the expression. If there was no 1 sitting there than we would have magically lost the dy/dx term in the original expression.
• Is this still counted as implicit derivatives? I'm confused because I thought explicit derivatives was where y was isolated on one side. What's the difference between the two? • The difference is that we have y terms on both sides of the equation (as y is part of the argument of the cos function). Although we have y on its own on the left-hand side, this is not the equation for y as a function of x. Thus, implicit differentiation is called for.
• Can I treat dy/dx as a fraction?

Can I multiply both sides by dx/dy to cleat the dy/dx?

I am not sure why one would do this, I just want to know if dy/dx is now its own thing or is it still a fraction. • Hello
I multiply the two paranteses before I multiply with -sin.
-sin (5x-3y)(5-3dy/dx) = -sin25x + sin 15x*dy/dx + sin15y - 9ydy/dx
My answer would then be :
dy/dx = (-sin25x+sin15y)/(-sin15x+1+9y)
Is this correct? How come Sal uses -5 sin (5x) and not -sin25x.? • No. The expression sin(5x-3y) means that you are taking the sine function of (5x-3y), That is not a multiplication and you certainly don't use the distributive property.
Thus, sin (kx) is NOT sine times kx. sin (kx) = 1/2 {ie^(-kxi)- ie^(kxi)}

Instead, you treat sin(5x-3y) as a single entity. You can break that up using a trigonometric identity, but that is not a more simple form.
For reference sake, though:
sin(5x-3y)=cos(3y)sin(5x)-cos(5x)sin(3y)

Finally, remember that k( sin x) is NOT equal to sin kx.
• At he substracted 3sin(5x-3y) from 1 and at the end he got -5sin(5x-3y) / (1-3sin(5x-3y)). However if we substract the other way and get 3sin(5x-3y)-1 we can get 5sin(5x-3y) / (3sin(5x-3y)-1) . Would it not be right as well? • At first, I got confused. However, I came to this conclusion: `d/dx` is the differential operator (similar to `+, -, *, /`). `dy/dx = d/dx (y)` is the variable (similar to `a = 5, b = 100, c = -0.12`); so we can apply operators to it, right? • I was just curious as to why we must again solve for dy/dx, since that is already what we have once we use the chain rule
(1 vote) • The problem is that you had dy/dx on both sides of the equation, and the goal was to find the derivative of y with respect to x. You need the dy/dx isolated for the same reason you don't leave a linear equation as y=2x-y. It makes it much simpler to do any follow up work if you needed the equation if it's already prepared for you. This way when you plug in values for your variables you don't need to do any more algebra, it's just simple order of operations. I hope that clarified everything a little for you.
• How to know that the equation given to us will be solved by applying product rule or chain rule ?
Please explain with the examples if possible .
Thank you .   