1. Total Differentiation; total mess.

Hello all,

A question about total differentiation is doing my head in. It is:

u = x^5 - 10x^3y^2 + 5xy^4 and y = 1 + x^2

It wants me to show that du/dx is some expression in x involving x's up to the power 8. But my technique gives rise to x's up to the power 12. I've been using du = (partial du/dx)dx + (partial du/dy)dy, and dividing through by dx, leaving du/dx on the left and then multiples of dx/dx (=1) and dy/dx (=2x) on the right.

Am I doing this wrong? Any help appreciated.

2. Are you sure you copied the right problem?

$\frac{du}{dx} = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} \frac{dy}{dx}$

$= 5x^{4} -30x^{2}y^{2}+5y^{4} + (-20x^{3}y + 20xy^{3})(2x)$

$= 5x^{4} -30x^{2}y^{2} + 5y^{4} -40x^{4}y + 40 x^{2}y^{3}$

3. You seem to have arrived at the same thing as me - the last term would appear to yield a term in x^12 when the substitution for y is used. This is a copy paste from the actual question on the website (i'd link you but you need a university account to get in):

Given that u = x5 -10x3y2 + 5xy4, and y = 1 + x2, show that
du
dx
= 5 + 30x2 + 55x4 + 70x6 + 45x8

The formatting isn't perfect but you can see what they think the answer is.

4. Never mind - I think every time I was taught about indices must have been a friday.

But anyway, assuming it does all work out, am I using the correct method?

5. Yeah, I forgot to substitute for y.

$5x^{4} -30x^{2}y^{2} + 5y^{4} -40x^{4}y + 40 x^{2}y^{3}$

$= 5x^{4} -30x^{2} (1+2x^{2}+x^{4}) + 5 (1+4x^{2}+6x^{4} +40x^{6} +x^{8}) -4x^{4}$ $(1+x^{2}) + 40x^{2}(1+3x^{2}+3x^{4}+x^{6})$

$= 5 + 30x^{2} + +55x^{4} +70x^{6}+ 45x^{8}$

In general, if $f(t) = f\big(x_{1}(t),x_{2}(t),...,x_{n}(t)\big)$

then $\frac{df}{dt} = \frac{\partial f}{\partial x_{1}} \frac{dx_{1}}{dt}+ \frac{\partial f}{\partial x_{2}} \frac{dx_{2}}{dt} + ... + \frac{\partial f}{\partial x_{n}} \frac{dx_{n}}{dt}$

In this particular problem we have $u(x) = u\big(x,y(x)\big)$

so $\frac{du}{dx} = \frac{\partial u}{\partial x} \frac{dx}{dx}+ \frac{\partial u}{\partial y} \frac{dy}{dx} = \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} \frac{dy}{dx}$