Problems & Solutions: Partial Differentiation 001

The first example is from: Widder, David V., "Advanced Calculus", Dover Books on Mathematics

Prerequisites: single-variable calculus, specifically u-substitution

Example 1:

Find $$\frac{\partial }{\partial x}\frac{sin(xy)}{cos(x+y)}$$

Solution:

We will need the quotient rule: $$\left ( \frac{f}{g} \right )' = \frac{f(x)g'(x) - f'(x)g(x)}{[g(x)]^{2}}$$

Setting $$f(x) = sin(xy)$$ and $$g(x) = cos(x+y)$$ we will need to find the derivatives of each.

Using the rule: $$\frac{\mathrm{d} }{\mathrm{d} x}sin(u(x)) = cos(u(x))\frac{\mathrm{d} u}{\mathrm{d} x}$$ $$f'(x) = cos(xy)\frac{\partial }{\partial x}(xy)$$ $$f'(x) = ycos(xy)$$

And conversely, setting $$g(x) = cos(x+y)$$

Using the rule: $$\frac{\mathrm{d} }{\mathrm{d} x}cos(u(x)) = -sin(u(x))\frac{\mathrm{d} u}{\mathrm{d} x}$$ $$g'(x) = -sin(x+y)\frac{\partial }{\partial x}(x+y)$$ $$g'(x) = -sin(x+y)(1+0)$$ $$g'(x) = -sin(x+y)$$

Putting the parts together: $$\frac{\partial }{\partial x}\frac{sin(xy)}{cos(x+y)} = \frac{sin(xy)(-sin(x+y))-(ycos(xy)cos(x+y))}{[cos(x+y)]^{2}}$$

This is usually sufficient for most professors. It is assumed that if you can get this far, you can simplify.