1. ## Implicit Differentiation

I am learning about implicit differentiation and although I understand the rules, I am having hard time understanding the notation.

I have failed to understand differentiation with regards the variables i.e. differentiation of a variable with respect to x and y. Lets start with a simple example. I will carry the calculations up to the point where I could ask all my questions:

(1) 7y4+x3y+x=4
(2) d/dx[7y4+x3y+x]=d/dx[4]
(3) d/dx[7y4]+d/dx[x3y]+d/dx[x]=0

We are at the point where I could ask my first question. Lets isolate the term to the right of the equal sign and assume that it is a separate function y=4 then dy/dx would be 0. What would the derivative be with respect to y? Continuing:

(4) 28y3[dy/dx]+x3[dy/dx]+3yx2+1=0

I think here we first assume that y=u and then take du/dx, am I correct? But then why do we continue to carry dy/dx? If we isolate d/dx[7y4] from (3) above, what would be its derivative with regards to y?

Please, explain as if you were explaining it to a four year old...

2. ## Re: Implicit Differentiation

Originally Posted by Soundork
I am learning about implicit differentiation and although I understand the rules, I am having hard time understanding the notation.

I have failed to understand differentiation with regards the variables i.e. differentiation of a variable with respect to x and y. Lets start with a simple example. I will carry the calculations up to the point where I could ask all my questions:

(1) 7y4+x3y+x=4
(2) d/dx[7y4+x3y+x]=d/dx[4]
(3) d/dx[7y4]+d/dx[x3y]+d/dx[x]=0 So far so good.

We are at the point where I could ask my first question. Lets isolate the term to the right of the equal sign and assume that it is a separate function y=4 then dy/dx would be 0. What would the derivative be with respect to y? Continuing:

(4) 28y3[dy/dx]+x3[dy/dx]+3yx2+1=0

This is correct … here's why (taking each term separately):

d/dx[7y4] = d/dy [7y^4] * [dy/dx] …. that's the chain rule
=28y^3 * [dy/dx]

d/dx [x^3y] … this is a product so you need to use the product rule ...let u= x^3 and v=y
=uv' + vu' (the derivatives are wrt x)
=x^3*d/dx(y) + y*d/dx(x^3)
=x^3 * [dy/dx] + y*3x^2
= x^3 [dy/dx] + 3x^2y

The third term is obvious.

I think here we first assume that y=u and then take du/dx, am I correct? No But then why do we continue to carry dy/dx? If we isolate d/dx[7y4] from (3) above, what would be its derivative with regards to y? In the second term you can assume that u= 7y^4, so you need to find du/dx = du/dy * dy/dx (Chain rule) That's why the dy/dx is still there.

Please, explain as if you were explaining it to a four year old...

(1)$7y^4+x^3y+x=4$
(1)$7y^4+x^3y+x=4$
$28y^3y'+[(3x^2y)+x^3y']+1=0$ Solve for $y'$.