Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6493

Exercise

Given the differentiable function (with parameters a,b)

$w=f(x+at, y+bt)$

Prove the equation

$w'_t=aw'_x+bw'_y$

Proof

Define

$u=x+at$

$v=y+bt$

We get the function

$w=f(u,v)$

And the internal functions

$u(x,t)=x+at$

$v(y,t)=y+bt$

We will use the chain rule to calculate the partial derivatives of w.

$w'_t=f'_u\cdot u'_t+f'_v\cdot v'_t=$

$=f'_u\cdot a+f'_v\cdot b$

$w'_x=f'_u\cdot u'_x+f'_v\cdot v'_x=$

$=f'_u\cdot 1+f'_v\cdot 0=$

$=f'_u$

$w'_y=f'_u\cdot u'_y+f'_v\cdot v'_y=$

$=f'_u\cdot 0+f'_v\cdot 1=$

$=f'_v$

We got the equations

$w'_t=f'_u\cdot a+f'_v\cdot b$

$w'_x=f'_u$

$w'_y=f'_v$

Hence, we get

$w'_t=w'_x\cdot a+w'_y\cdot b$

$w'_t=aw'_x+bw'_y$

As required.

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