5.7 Partial derivatives, differentiation and the gradient

Our next goal is to extend the notion of differentiability of functions in one variable \(f:\R \to \R\) to functions in two variables \(f:\R^2 \to \R\).

Let us recall first the notion of differentiability

Let \(f: (a, b) \to \R\) be a function. We say that \(f\) is differentiable at \(x_0 \in (a, b)\) if the following limit exists \(\lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h} = \ell\). In such a case we write \(f'(x_0) = \ell\).

Another way of writing this limit is \(\ell = \lim_{x \to x_0} \frac{f(x) - f(x_0)}{x - x_0}\)

Note that if such a limit exists, then \(\lim_{x \to x_0} \frac{f(x) - (f(x_0) + \ell(x - x_0))}{x - x_0} = 0\)

So, one may say that the function \(f\) is differentiable at \(x_0 \in (a, b)\) if there exists \(\ell \in \R\) such that the limit above is 0.

The function: \(y = f(x_0) + f'(x_0)(x - x_0)\) is the tangent line of \(f\) at \(x_0\).

Here we can approximate \(f(x)\) by a tangent line

When trying to understand what happens to a function on two variables, one may try to look at one variable at the time.

Example

Let's analyze what happens at \(p=(0,0)\) where \(f(x,y)=x^2+y^2\) and \(y=0\). We are looking at the intersection of these two

Then we want the intersection of \((x,y,x^2+y^2)\) and \((x,0,z)\), that is \((x,0,x^2)\), that is \(f(x,0)=g(x)=x^2\)

The intersection formula is differentiable since \(g'(0)=\lim_{x\to 0}\frac{g(x)-g(0)}{x-0}=\lim_{x\to0}\frac{x^2}{x}=0\Rightarrow g'(0)=0\)

Then \(y=g(0)+g'(0)(x-0)=0\)

Actually, we can do this for any \(x_0\), \(g'(x_{0})=\lim_{x\to x_0}\frac{g(x)-g(x_0)}{x-x_0}=\lim_{x\to x_0}\frac{x^2-x_0^2}{x-x_0}=2x_0\Rightarrow g'(x_0)=2x_0\)

As \(x_0\) was arbitrary, we get a function \(g'(x)=2x\)

For \(f(x,y)=x^2+y^2\) and \(x=0\), the intersection function is \((0,y,y^2)\), that is \(f(0,y)=g(y)=y^{2}\)

Same process, we get \(g'(y)=2y\)

Then we have \(y=0\), then we have \((x,0,0)\) and \((0,y,0)\), this is \(z=0\) plane, that is the derivative of \((0,0)\)

Questions:

Does this mean that our function is differentiable at \((0,0)\)?
We have at least two tangent lines. Are all possible tangent lines in the same plane?
Our intuition says that a function \(f(x, y)\) is differentiable at \((x_0, y_0)\) if there is a tangent plane at the point \((x_0, y_0, f(x_0, y_0))\) on the graph of \(f(x, y)\) in \(\R^3\).

How do we formalize this notion?

Let us start by formalizing the derivatives with respect to one variable:

Partial derivatives

Definitions:

The partial derivative of a function \(f:\R^2 \to \R\) with respect to the variable \(x\) at the point \((x_0, y_0)\) is the limit

\[ \ell=\lim_{x\to x_0}\frac{f(x,y_{0})-f(x_{0},y_{0})}{x-x_{0}} \]

If it exists, we write: \(\ell = \frac{\partial f}{\partial x}(x_0, y_0)\)

The partial derivative of a function \(f:\R^2 \to \R\) with respect to the variable \(y\) at the point \((x_0, y_0)\) is the limit

\[ \ell=\lim_{y\to y_0}\frac{f(x_{0},y)-f(x_{0},y_{0})}{y-y_{0}} \]

If it exists, we write: \(\ell = \frac{\partial f}{\partial y}(x_0, y_0)\)

Example

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

Let \(p = (0,0) = (x_0, y_0)\)

Then \(\frac{\partial f}{\partial x}(0,0) = 0 \quad , \quad \frac{\partial f}{\partial y}(0,0) = 0\)

When taking \(y = y_0\), we are fixing the variable.
For any point, we may compute \(\frac{\partial f}{\partial x}(x,y)\) by considering \(y\) as a number (scalar)

Then \(\frac{\partial f}{\partial x}(x, y) = 2x \quad , \quad \frac{\partial f}{\partial y}(x, y) = 2y\)
\(f(x,y)=x^2-y^2\)

\(p=(x_{0},y_{0})=(0,0)\), then \(\frac{\partial f}{\partial x}(x_0,y_0)=\lim_{x\to x_0}\frac{f(x,y_0)-f(x_0,y_0)}{x-x_0}=\lim_{x\to x_0}\frac{x^2-y_0^2-x_0^2+y_0^2}{x-x_0}=\lim_{x\to x_0}\frac{x^2-x_0^2}{x-x_0}=2x_0\)

Then \(\frac{\partial f}{\partial x}(x,y)=2x,\frac{\partial f}{\partial x}(0,0)=0\)

\(\frac{\partial f}{\partial y}(x_{0},y_{0})=\lim_{y\to y_0}\frac{f(x_{0},y)-f(x_{0},y_{0})}{y-y_{0}} =\lim_{y\to y_0}\frac{x_{0}^{2}-y^{2}-x_{0}^{2}+y_{0}^{2}}{y-y_{0}}=\lim_{y\to y_0} \frac{y_{0}^{2}-y^{2}}{y-y_{0}}=-2y_{0}\)

Then \(\frac{\partial f}{\partial y}(x,y)=-2y,\frac{\partial f}{\partial y}(0,0)=0\)
\(f(x,y)=\begin{cases} \frac{xy}{x^2+y^2} \quad \text{ if }(x,y)\neq (0,0)\\ 0\quad \quad \text{ if } (x,y)= (0,0) \end{cases}\)

Line \(y=0,p=(0,0)\), then \(f(x,0)=0\), then \(\frac{\partial f}{\partial x}(0,0)=0\)

Line \(x=0,p=(0,0)\), then \(f(0,y)=0\), then \(\frac{\partial f}{\partial y}(0,0)=0\), thus then plane is \(z=0\)

But we proved that this function is not continuous at \(p=(0,0)\), so it should not be differentiable

The linear or affine approximation

The geometrical idea behind differentiation is that a function is differentiable at a point \((x_0, y_0)\) if the function can be approximated by a linear (or affine) function at that point.

For functions \(f:\R \to \R\), the linear function is the tangent line. In case of functions \(f:\R^2 \to \R\), the tangent line should be replaced by a tangent plane.

The general equation of a non-vertical plane passing through a point \((x_0, y_0, f(x_0, y_0))\) is:

\(z = a(x - x_0) + b(y - y_0) + f(x_0, y_0)\) with \(a, b \in \R\).

This plane gives a good approximation of the function at \((x_0, y_0)\) if \(\lim_{(x,y) \to (x_0, y_0)} \frac{f(x, y) - (a(x - x_0) + b(y - y_0) + f(x_0, y_0))}{\| (x, y) - (x_0, y_0) \|} = 0\)

Note that here we are using the distance in \(\R^2\)

Let us determine the constants \(a, b \in \R\). We assume the limit exists, then approach the point \((x_0,y_0)\) using the line \(y=y_0\), we should get the same limit

\(0=\lim_{(x,y_0)\to(x_0,y_0)}\frac{f(x,y_{0})-(a(x-x_{0})+b(y_{0}-y_{0})+f(x_{0},y_{0}))}{\sqrt{\left(x-x_{0}\right)^{2}+\left(y_{0}-y_{0}\right)^{2}}} =\lim_{x\to x_0}\frac{f(x,y_{0})-\left(a(x-x_{0})+f\left(x_{0},y_{0}\right)\right)}{x-x_{0}}\)

\(=\lim_{x\to x_0^{+}}\frac{f(x,y_{0})-f\left(x_{0},y_{0}\right)-a(x-x_{0})}{x-x_{0}} =\lim_{x\to x_0^{+}}\frac{f(x,y_{0})-f\left(x_{0},y_{0}\right)}{x-x_{0}}-a\)

Then \(a=\lim_{x\to x_0^{+}}\frac{f(x,y_{0})-f\left(x_{0},y_{0}\right)}{x-x_{0}}=\frac{\partial f}{\partial x}\left(x_{0},y_{0}\right)\)

Analogously, \(b=\frac{\partial f}{\partial y}(x_0,y_0)\)

Definition

Differentiable

Let \(f:\R^2 \to \R\) be a function. We say that \(f\) is differentiable at \((x_0, y_0)\) if both partial derivatives exist at that point and

\[ \lim\limits_{(x,y) \to (x_0, y_0)}\frac{f(x, y) - \left( \frac{\partial f}{\partial x}(x_{0}, y_{0})(x - x_{0}) + \frac{\partial f}{\partial y}(x_{0}, y_{0})(y - y_{0}) + f(x_{0}, y_{0}) \right)}{\|(x, y) - (x_{0}, y_{0})\|}= 0 \]

Tangent Plane

Let \(f:\R^2 \to \R\) be differentiable at \((x_0, y_0)\). Then the plane

\[ z=f(x_{0},y_{0})+\frac{\partial f}{\partial x}(x_{0},y_{0})(x-x_{0})+\frac{\partial f}{\partial y}(x_{0},y_{0})(y-y_{0}) \]

is called the tangent plane of the graph of \(f\) at the point \((x_0, y_0, f(x_0, y_0))\).

Gradient Vector

If both partial derivatives of \(f\) exist at a point \(p = (x_0, y_0)\), the vector in \(\R^2\):

\[ \nabla f(x_{0},y_{0})=\left(\frac{\partial f}{\partial x}(x_{0},y_{0}),\frac{\partial f}{\partial y}(x_{0},y_{0})\right) \]

is called the gradient of \(f\) at \(p\).

Example

\(f(x,y)=x^2+y^2\) is differentiable for all \((x,y)\in \R^2\)

But let's check it at \(p=(0,0)\), then \(\frac{\partial f}{\partial x}(0,0)=0,\frac{\partial f}{\partial y}(0,0)=0\)

Then \(\lim_{(x,y)\to(0,0)}\frac{f(x,y)-\left(0(x-0)+0(y-0)+f(0,0)\right)}{\sqrt{(x-0)^{2}+(y-0)^{2}}} =\lim_{(x,y)\to(0,0)}\frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}}}=\lim_{(x,y)\to(0,0)}\sqrt{x^{2}+y^{2}} =0\)

Then tangent plane is \(z=f(0,0)+0(x-0)+0(y-0)=0\)
\(f(x,y)=x^{2}-y^{2}\)

At \(p=(0,0)\), then \(\frac{\partial f}{\partial x}(0,0)=0,\frac{\partial f}{\partial y}(0,0)=0\)

Then \(\lim_{(x,y)\to(0,0)}\frac{f(x,y)-\left(0(x-0)+0(y-0)+f(0,0)\right)}{\sqrt{(x-0)^{2}+(y-0)^{2}}} =\lim_{(x,y)\to(0,0)}\frac{x^{2}-y^{2}}{\sqrt{x^{2}+y^{2}}}=\lim_{(x,y)\to(0,0)}\frac{x^{2}-y^{2}}{\sqrt{x^{2}+y^{2}}}\)

Then use comparison test, since \(0\leq \frac{x^{2}}{\sqrt{x^2+y^2}}\leq \frac{x^2+y^2}{\sqrt{x^2+y^2}}\) and \(\lim_{(x,y)\to(0,0)}\frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}}}=0\)

Then \(\lim_{(x,y)\to(0,0)}\frac{x^{2}-y^{2}}{\sqrt{x^{2}+y^{2}}}=0\)

Then tangent plane is \(z=f(0,0)+0(x-0)+0(y-0)=0\)
\(f(x,y)= \begin{cases} \frac{xy}{x^{2}+y^{2}} \quad \text{ if }(x,y)\neq (0,0) \\ 0\quad \quad \text{ if } (x,y)= (0,0) \end{cases}\)

At \(p=(0,0)\), then \(\frac{\partial f}{\partial x}(0,0)=0,\frac{\partial f}{\partial y}(0,0)=0\)

Then \(\lim_{(x,y)\to(0,0)}\frac{f(x,y)-\left(0(x-0)+0(y-0)+f(0,0)\right)}{\sqrt{(x-0)^{2}+(y-0)^{2}}} =\lim_{(x,y)\to(0,0)}\frac{\frac{xy}{x^{2}+y^{2}}}{\sqrt{x^{2}+y^{2}}}=\lim_{(x,y)\to(0,0)} \frac{xy}{\left(x^{2}+y^{2}\right)^{\frac{3}{2}}}=\nexists\)

We can take \(y=x\), then \(\lim_{(x,y)\to(0,0)}\frac{x^2}{\left(2x^{2}\right)^{\frac{3}{2}}}=\infty\)

Thus it is not differentiable at \((0,0)\)

Remarks

Using the dot product, the equation of the tangent plane can be written as:

\(z = f(x_0, y_0) + \nabla f(x_0, y_0) \cdot (x - x_0, y - y_0)\)

Or more compactly, if \(p = (x_0, y_0)\) and \(x = (x, y)\): \(z = f(p) + \nabla f(p) \cdot (x - p)\)
If \(f:\R^2 \to \R\) is differentiable at \(p = (x_0, y_0)\), then the vector: \(\left( \frac{\partial f}{\partial x}(x_0, y_0), \frac{\partial f}{\partial y}(x_0, y_0), -1 \right)\)is orthogonal to the tangent plane at \(p\).