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### Course: Multivariable calculus>Unit 2

Lesson 1: Partial derivatives

# Partial derivatives, introduction

Partial derivatives tell you how a multivariable function changes as you tweak just one of the variables in its input. Created by Grant Sanderson.

## Want to join the conversation?

• at and , it is shown that the derivative of sin(y) = 0. I understand that the derivative of a constant is 0, and sin(y) is a constant when evaluating df/dx, but the derivative of sin is also cosine. Why wouldn't it be cos(2) (or cos(y)) instead of 0?
• 3blue1brown
• Around the 6 to 7 minute mark, how does squaring X and Y treat it as a constant?
• When you are taking the partial derivative with respect to x, you treat the variable y as if it is a constant. It is as if you plugged in the value for y ahead of time. This means an expression like y^2 just looks like (some constant)^2, which is again a constant.

For example, if ultimately you plan to plug in y=5, when you see an expression like y^2, you would treat it as 5^2 = 25, which is again a constant.
• at , why is it sin (y) and not cos(y)
• did you find an answer cause GPT agrees haha!
(1 vote)
• So what is then the full derivative of a multi-veriable function? Does it extend backwards to R2 and R1 and to higher dimensions?
• There is no "full derivative". The closest thing would be the gradient, which I assume is covered in later sections.
• About 7 minutes in he says the derivative of 1^2*y is 1. Is this not the same as y^2 of which the derivative should be 2y?
• 1^2 * y isn't equal to y^2 unless im missing something here
(1 vote)
• You sound just like 3blue1brown!
• he is the same guy
• At , when you're talking about partial derivatives, are you saying that you have to look at changes in "x" and "y" separately? You can't see what that does combined? Is that where directional derivatives come in?
• In single variable calculus, excluding implicit differentiation, the derivative of a function, f(x), equals dy/dx, and the x variables have their own derivative notation, dx/dx, but that cancels out to one which is why it is not commonly written out. So maybe you could get df/dx and df/dy and have df/dx divided by df/dy which would cancel out df and give you dy/dx, the corresponding change between the variables as values differ. ¯\(ツ)/¯ but I don't know for sure, I just started learning yesterday to get an idea of multivariable calculus
• at you say Cos(1) is what we had before but surely you meant Cos(2)