Today I learnt about implicit differentiation using this: d/dx(f) = df/dy x dy/dx i don't understand when doing implicit differentiation how the d/dy part works for y terms: d(y^2)/dx=d(y^2)/dy x dy/dx=2y dy/dx Why is the derivative of y^2 with respect to y, 2y? Do you assume it's a function of something else? And similarly if you apply this same formula to an x term (even though its not needed) you get: d(x^2)/dx=d(x^2)/dyx(dy)/(dx) How does that simplify to the 2x I know it is?

Kymani Hatfield 2022-10-19 Answered
Today I learnt about implicit differentiation using this:
d f d y × d y d x
I don't understand when doing implicit differentiation how the d/dy part works for y terms:
d ( y 2 ) d x = d ( y 2 ) d y × d y d x = 2 y d y d x
Why is the derivative of y2 with respect to y, 2 y? Do you assume it's a function of something else? And similarly if you apply this same formula to an x term (even though its not needed) you get:
d ( x 2 ) d x = d ( x 2 ) d y × d y d x
How does that simplify to the 2 x I know it is?
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Answers (2)

Gael Irwin
Answered 2022-10-20 Author has 13 answers
Yes, here they assume that y is a function of something else.

Whenever we apply implicit differentiation like with d y 2 ( x ) d x , we sort of first forget it is a function of something else, say x, we just derive it using the chain rule and then we need to correct for this via the familiar additional factor:
d y 2 ( x ) d x = 2 y ( x ) d y ( x ) d x
Actually implict differentiation often boils down to just using the chain rule. Notice that if y is not a function of x, we get that d y d x = 0 and the equation just reads 0=0.

For your second case where you state:
d ( x 2 ) d x = d ( x 2 ) d y × d y d x
We could call y = x 2 because you really need to tell the reader what this new variable y is in relation to the old variable x, else the derivative could be anything. We then get:
d y d x = d y d y × d x 2 d x = 1 2 x
In general we use implicit differentiation when there is some complicated variable dependence and you cannot directly express some variable explicitly in terms of the other. An example of such an equation would be:
t + t 2 + 3 z 2 + z 3 = 6
Using implicit differentiation we could construct the tangent line for ( t , z ) = ( 1 , 1 ), we will derive both sides with respect to t, we do not know if z is is a function of t so we need to use the chain rule.

We would get:
1 + 2 t + 6 z d z d t + 3 z 2 d z d t = 0
Now we plug in the value at (1,1) to get:
1 + 2 + 6 d z d t + 3 d z d t = 0
Or after rearranging:
9 d z d t = 3 d z d t = 1 3
We get the tangent line z = 1 3 ( t 1 ) + 1
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propappeale00
Answered 2022-10-21 Author has 5 answers
You implicitly utilize the formula of derivative of a composite functions or a chain rule:
f = y ( x ) 2 , d f d x = d ( y 2 ) d x , f = g ( y ( x ) ) , g ( y ) = y 2 , d d x g ( y ( x ) ) = d g ( y ( x ) ) d y d y ( x ) d x = ( 2 y ( x ) ) d y d x
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