# Partial derivative - problem with understanding

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

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.

• Great thanks, it of course works, but I still can't figure out how you received $y-9=6(x-3)$. What is this? It looks like it's neither $f(x)$ nor $f'(x)$. It looks like some mix of them. Jul 16, 2018 at 10:21
• The same as in José Carlos Santos' answer: the line through $(a,b)$ with slope $m$ is given by $y-b=m(x-a)$; here $b=f(a)$ and $m=f'(a)$, the derivative. Jul 16, 2018 at 11:23

Since you are dealing with functions of one variable, your problem is not about partial derivatives. It's just about derivatives.

Concerning your specific example ($$y=x^2$$), the derivative is $$2x$$, yes. What that means is that the slope of the tangent line of the graph of $$y=x^2$$ that passes through $$(x_0,{x_0}^2)$$ is $$2x_0$$. And you can check that, indeed, the line $$y=2x_0(x-x_0)+{x_0}^2(=2x_0x-{x_0}^2)$$ is indeed tangente to the graph of that function. See the image below. • Ok, great thanks, it works, but I still can't figure out how you've found that $y = 2x_0 (x-x_0) + (x_0)^2$ It makes no sense for me. Is that some formula? Jul 16, 2018 at 10:49
• The only line with slope $m$ passing through the point $(a,b)$ is $y=m(x-a)+b$. I applied this formula with $a=x_0$, $b={x_0}^2$, and $m=2x_0$. Jul 16, 2018 at 10:52
• :) now it seems to be obvious, thanks for your patience. Jul 16, 2018 at 12:10