11. Partial Derivatives and Tangent Planes

b. Algebraic Definition of Partial Derivatives

For a function of one variable, \(f(x)\), the definition of the derivative is: \[ f'(x)=\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x)}{h} \] For a function of two variables, \(f(x,y)\), the partial derivatives are computed by differentiating with respect to one variable while holding the other variable fixed. So the formulas for the partial derivatives are:

The partial derivatives of \(f(x,y)\) are \[ f_x(x,y)=\lim_{h\rightarrow 0} \dfrac{f(x+h,y)-f(x,y)}{h} \] \[ f_y(x,y)=\lim_{h\rightarrow 0} \dfrac{f(x,y+h)-f(x,y)}{h} \]

Looking at the formulas, the \(x\)-partial derivative says how the function changes as we move parallel to the \(x\)-axis, that is in the \(\hat{\imath}\) direction. Similarly, the \(y\)-partial derivative says how the function changes as we move parallel to the \(y\)-axis, that is in the \(\hat{\jmath}\) direction.

More directions exist. In fact, we can take a derivative along any vector \(\vec v\) .

Given any vector \(\vec v\), the derivative of \(f(x,y)\) along \(\vec v\) (or with respect to \(\vec v\)) is: \[ \nabla_{\vec v}f(x,y) =\lim_{h\rightarrow 0} \dfrac{f((x,y)+h\vec v)-f(x,y)}{h} \]

The symbol \(\nabla\) is read as "del". It is an upside down, capital Greek delta: \(\Delta\). If you want, you can think of \(\nabla\) as an upper case partial \(\partial\).

In particular, the \(x\)-partial derivative is the derivative along \(\hat{\imath}\) since: \[ (x,y)+h\hat{\imath}=(x+h,y) \] and the \(y\)-partial derivative is the derivative along \(\hat{\jmath}\) since: \[ (x,y)+h\hat{\jmath}=(x,y+h) \] So: \[ f_x(x,y)=\nabla_{\hat{\imath}}f(x,y) \qquad \text{and} \qquad f_y(x,y)=\nabla_{\hat{\jmath}}f(x,y) \]

Once we cover the chain rule, we will be able to give a formula for the derivative along a vector that does not involve limits.

These limit formulas for partial derivatives and derivatives along a vector are rarely use in computations. They are mostly used in theoretical derivations. However, they can be turned around and used to compute a limit:

Compute \(\displaystyle \lim_{h\rightarrow 0} \dfrac{(y+h)\sin(xy+xh)-y\sin(xy)}{h}\).

\(\displaystyle \lim_{h\rightarrow 0} \dfrac{(y+h)\sin(xy+xh)-y\sin(xy)}{h} =\dfrac{\partial}{\partial y}(y\sin(xy))\)

\(\displaystyle \lim_{h\rightarrow 0} \dfrac{(y+h)\sin(xy+xh)-y\sin(xy)}{h} =xy\cos(xy)+\sin(xy)\)

Since there is a \((y+h)\) in the numerator, we suspect this is a \(y\)-partial derivative. So we compare \[ \lim_{h\rightarrow 0} \dfrac{(y+h)\sin(xy+xh)-y\sin(xy)}{h} \] to \[ f_y(x,y)=\lim_{h\rightarrow 0} \dfrac{f(x,y+h)-f(x,y)}{h} \] From the second term in the numerator, we identify: \[ f(x,y)=y\sin(xy) \] We then check: \[ f(x,y+h)=(y+h)\sin(x[y+h])=(y+h)\sin(xy+xh) \] which is the first term in the numerator. So: \[\begin{aligned} \lim_{h\rightarrow 0} &\dfrac{(y+h)\sin(xy+xh)-y\sin(xy)}{h} =\dfrac{\partial}{\partial y}(y\sin(xy)) \\ &=y\cos(xy)x+\sin(xy) =xy\cos(xy)+\sin(xy) \end{aligned}\]

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