sorry for such basic question but I have problem with understanding partial derivative.
I watched some videos about that, and read some articles, bo with no result.
As I understand partial derivative gives me functions which are tengent to my main (parent) function.
In all tutorials that I've seen, they give example for $f(x,y)$. What you can write as $z = f(x,y)$
But I need tangent line to function of just one variable, like $f(x)$
So it seems to be obvious to just set $z=0$
So let's say my function is $f(x) = x^2$
I can write it as $y = x^2$
So after partial derivative on $x$, I get:
$y=2x$
And it's definetaly not tangent to $y=x^2$
I see I make some logical error, but don't know where and how to fix it.
Could anyone help me please?
Great thanks in advance
Best Answer
Your problem is not with partial derivatives, but with the interpretation of the (simple, one variable) derivative. Carefully read through the following:
The derivative of a function $f$ at a point $x=a$ is the slope of the tangent line to the graph of $f$ at the point $(a,f(a))$.
It's important to realize that the derivative at a point is a number, not a/the tangent line.
Let's take your example of the quadratic function of one variable: $$f(x)=x^2$$ The derivative is a function too and to avoid confusion I wouldn't write $y=2x$ for it but $f'(x)=2x$ or, if you are using $y=x^2$, then $y'=2x$.
Now if you want a tangent line, you need it somewhere on the graph of $f$, i.e. you need to pick a point where you want to find and/or draw the tangent. Say you want the tangent at $x=3$, where $y=3^2=9$, then the derivative gives you the desired slope: $f'(3)=2\cdot 3=6$.
The tangent line at $(3,9)$ is then given by $y-9=6\left(x-3\right) \iff y=6x-9$.
In summary: $y'=2x$ is not tangent to $y=x^2$, but the slope of the tangent to $y=x^2$ is given by $y'=2x$, evaluated where you want the tangent line.