In one line
For \(u=f(x,y)\), the total differential \(du=\dfrac{\partial u}{\partial x}dx+\dfrac{\partial u}{\partial y}dy\) gives the approximate change in \(u\) when \(x\) and \(y\) change by small amounts \(dx\) and \(dy\).
Explained simply
If your marks depend on two tests and both change a little, your total changes by (effect of test 1) + (effect of test 2). The partial derivatives tell you how sensitive \(u\) is to each input, and the total differential adds up the effects. It is used in labs to estimate how measurement errors in length, radius or time affect a calculated result.
Technically, if \(u=f(x,y)\) has continuous partial derivatives, then for small \(\Delta x,\Delta y\), \(\Delta u\approx du=u_x\,dx+u_y\,dy\). For three variables add \(u_z\,dz\). This is a self-learning topic in the syllabus.
How it works
- Write the formula for the quantity, e.g. area \(A=xy\).
- Find the partial derivatives at the given point.
- Multiply each by the given small change and add.
- For percentage error, take log first: \(\log A=\log x+\log y\) gives \(\dfrac{dA}{A}=\dfrac{dx}{x}+\dfrac{dy}{y}\), then multiply by 100.
Key formula
\(du=u_xdx+u_ydy\) (two variables), \(du=u_xdx+u_ydy+u_zdz\) (three variables). \(dx,dy\) carry the units of \(x,y\); \(du\) carries the units of \(u\). Relative error \(=\dfrac{du}{u}\); percentage error \(=\dfrac{du}{u}\times100\).
Quick example
A rectangle measures \(x=10\) cm by \(y=5\) cm. Each side is increased: \(dx=0.1\) cm, \(dy=0.05\) cm. Area \(A=xy\).
\(dA=y\,dx+x\,dy=5(0.1)+10(0.05)=0.5+0.5=1\) cm².
Actual new area \(=10.1\times5.05=51.005\) cm², so the true change is \(1.005\) cm². The differential estimate of 1 cm² is very close.
Percentage form: \(\dfrac{dA}{A}\times100=\left(\dfrac{0.1}{10}+\dfrac{0.05}{5}\right)\times100=2\%\).
Common mistakes
- Using large changes; the approximation only works when \(dx,dy\) are small.
- Forgetting to evaluate the partial derivatives at the given point.
- Adding percentage errors without the powers: for \(V=\pi r^2h\), \(\dfrac{dV}{V}=2\dfrac{dr}{r}+\dfrac{dh}{h}\).
How MU asks it
As a self-learning topic it may appear as a short application part: "Find the percentage error in the area of a rectangle when an error of 1% is made in measuring each side" (answer 2%), or "If \(u=x^2y\), find \(du\)". Write the formula, the partial derivatives, the substitution with units, and a clear final value.