Messy about partial derivatives

Hi! That’s it, I’m a little confused about the partial derivatives calculus. Luis has described two different ways to get the partial derivatives. I mean: In the video “Partial derivatives - Part I”, when treating the other variable as a constant, the partial derivatives resulting doesn’t contain the fixed one:

f(x) = x^2 + y^2

Result given:
df over dx = 2x ——> Not =2x+y^2 (Dropping y^2)
df over dy = 2y ——> Not =x^2+2y (Dropping x^2)

However, in the second example from the video “Partial derivatives - Part II” gives another example:

f(x) = 3x^2y^3

Result given:
df over dx = 6x·y^3 ——> Not just =6x (Keeping y^3)
df over dy = 9x^2·y^2 ——-> Not just =9y^2 (Keeping x^2)

Why is he dropping out the fixed variable from the partial derivatives in the first example, but in the second one he is keeping the fixed variable?
I can’t figure it out…

Thank you!

You are confused about how constants are treated in different derivative rules. Read on to understand how y should be treated in your two different scenarios:

Consider the first example you gave:
f(x) = x^2 + y^2

[1] Differentiate both side respective to \delta x:
\frac{\delta}{\delta x} f(x) = \frac{\delta}{\delta x}(x^2 + y^2)

[2] Sum rule of derivative: the derivative of a sum is equal to the sum of the derivatives
\frac{\delta}{\delta x} f(x) = \frac{\delta}{\delta x}x^2 + \frac{\delta}{\delta x}y^2

[3] For the first term \frac{\delta}{\delta x}x^2, apply the Power Rule
\frac{\delta}{\delta x} f(x) = 2x + \frac{\delta}{\delta x}y^2

[4] For the second term \frac{\delta}{\delta x}y^2, since y^2 is the constant when differentiate respective to x, its value is 0.
\frac{\delta}{\delta x} f(x) = 2x + 0

Remember that \frac{\delta}{\delta x} C = 0 (which means the constant will never change when we change x). This is why people automatically “ignore” summation terms without x when do differentiation with respect to x.

[5] Simplify:
\frac{\delta}{\delta x} f(x) = 2x

Now consider the second example you gave:

f(x) = 3x^2y^3

[1] Again, differentiate both side with respective to \delta x
\frac{\delta}{\delta x} f(x) = \frac{\delta}{\delta x}(3x^2y^3)

[2] Since y^3 is constant when differentiate respective to x, we can rewrite:

\frac{\delta}{\delta x} f(x) = \frac{\delta}{\delta x}(3y^2 \times x^2)

I think you was confused that y^2 can be ignored here because it can be “ignored” above. This is not correct as the y^2 is part of the whole term 3x^2y^3 (which contains x). We should see that 3y^2 is in fact the whole constant.

[3] Apply the Constant Multiple Rule

\frac{\delta}{\delta x} f(x) = 3y^2\times\frac{\delta}{\delta x}(x^2)

[4] From there apply the Power Rule:

\frac{\delta}{\delta x} f(x) = 3y^2 \times 2 \frac{\delta}{\delta x}x

[5] Simplify:

\frac{\delta}{\delta x} f(x) = 6y^2\frac{\delta}{\delta x}x

Hello @David73
I am yet to watch the video but in addition to what @ngtrhieu has said, In Partial derivates, if the operation between the variables is Addition like in your first example you drop the fixed variable but if the operation is Multiplication you keep the fixed variable. Hope this helps

Happy Learning

Wow! Thank you so much!

Clear as water! That’s what I need to remember the process :slight_smile:

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