# Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6511

Exercise

Given the differentiable function (with parameter a)

$$y=f(x-at)+g(x+at)$$

Prove the equation

$$y''_{tt}=a^2y''_{xx}$$

Proof

Define

$$u=x-at$$

$$v=x+at$$

We get the function

$$y=f(u)+g(v)$$

And the internal functions

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

$$v(y,t)=x+at$$

We will use the chain rule to calculate the first and second order partial derivatives of y.

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

$$=f'_u\cdot (-a)+g'_v\cdot a$$

$$=-af'_u+ag'_v$$

$$y''_{tt}=-a\cdot f''_u\cdot u'_t+ag''_v\cdot v'_t=$$

$$=-af''_u\cdot (-a)+ag''_v\cdot a=$$

$$=a^2f''_u+a^2g''_v$$

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

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

$$=f'_u+g'_v$$

$$y''_{xx}=f''_u\cdot u'_x+g''_v\cdot v'_x$$

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

$$=f''_u+g''_v$$

We will put the second order partial derivatives in the left side of the equation we need to prove.

$$y''_{tt}=a^2f''_u+a^2g''_v=$$

$$=a^2(f''_u+g''_v)=$$

$$=a^2y''_{xx}$$

We got to the right side of the equation as required.

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