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


2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – pajczur
    Jul 16, 2018 at 10:21
  • $\begingroup$ 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. $\endgroup$
    – StackTD
    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.

enter image description here

  • $\begingroup$ 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? $\endgroup$
    – pajczur
    Jul 16, 2018 at 10:49
  • $\begingroup$ 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 $m=2x_0$. $\endgroup$ Jul 16, 2018 at 10:52
  • $\begingroup$ :) now it seems to be obvious, thanks for your patience. $\endgroup$
    – pajczur
    Jul 16, 2018 at 12:10

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