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# Implicit differentiation (advanced example)

## Video transcript

let's say we have the relationship Y is equal to cosine of 5x minus 3y and what I want to find is the rate at which Y is changing with respect to X and we'll assume that Y is a function of X so let's do what we've always been doing let's apply the derivative operator to both sides to both sides of this equation on the left hand side right over here we get dy/dx is equal to now here on the right hand side we're going to apply the chain rule the derivative of the cosine of something with respect to that something is going to be equal to negative sine negative sine of that something so negative sine of 5x minus 3y and we have to multiply that by the derivative of that something with respect to X so what's the derivative of the something with respect to X well the derivative of 5x with respect to X is just equal to 5 and the derivative of negative 3 y with respect to X is just negative 3 times dy/dx the negative 3 times the derivative of Y with respect to X and now we just need to solve for dy/dx as you can see with some of these implicit differentiation problems this is the hard part and actually let me make that dy/dx the same color so that we can keep track of it easier so this is going to be dy/dx and then I can close the parentheses so how can we do it it's just going to be a little bit of algebra to work through well we can distribute the sine of 5x minus 3y so let me rewrite everything we get dy whoops I'm going to do that into the yellow color we get dy/dx is equal to you distribute the negative sine of 5x minus 3y you get so let me make sure we know what we're doing it's going to be we're going to distribute that and we're going to distribute that you're going to 5 times all of this so you're going to have negative this 5 times the sine sine of 5x minus 3y and then you're going to have the negative times a negative those are going to you end up with a positive and so you're getting up with plus plus three times the sign three times the sine of 5x minus 3y dy DX dy DX D Y DX now what we can do is subtract three sine of 5x minus 3y from both sides so just to be clear this is essentially a 1 dy/dx so if we subtract this from both sides we are left with so on the left hand side we're going to have a 1 dy DX and we're going to subtract from that 3 sine of 5x minus 3y dy DX's you're going to have 1 minus 3 I'll keep the color for the 3 for fun 3 sine of 5x minus 3y dy DX's on the left hand side on the left hand side is going to be equal to is going to be equal to well we subtracted this from both sides so on the right hand side this is going to go away so we're just going to be left with a negative 5 sine of sine of 5x minus 3y and we're in the homestretch now to solve for dy/dx we just have to divide both sides of the equation by this and we are left with dy/dx is equal to this thing negative 5 times the sine of sine of 5x minus 3y all of that over 1 minus 3 3 sine sine of 5x minus 3y and we are done