# 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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