75. We take the derivative with respect to x of both sides of Eq. 34-11:

∂

∂x

∂E

∂x



=

∂

2

E

∂x

2

=

∂

∂x

−

∂B

∂t



=

−

∂

2

B

∂x∂t

.

Now we differentiate both sides of Eq. 34-18with respect to t:

∂

∂t

−

∂B

∂x



=

−

∂

2

B

∂x∂t

=

∂

∂t

ε

0

µ

0

∂E

∂t



= ε

0

µ

0

∂

2

E

∂t

2

.

Substituting ∂

2

E/∂x

2

=

−∂

2

B/∂x∂t from the first equation above into the second one, we get

ε

0

µ

0

∂

2

E

∂t

2

=

∂

2

E

∂x

2

,

or

∂

2

E

∂t

2

=

1

ε

0

µ

0

∂

2

E

∂x

2

= c

2

∂

2

E

∂x

2

.

Similarly, we differentiate both sides of Eq. 34-11 with respect to t

∂

2

E

∂x∂t

=

−

∂

2

B

∂t

2

,

and differentiate both sides of Eq. 34-18with respect to x

−

∂

2

B

∂x

2

= ε

0

µ

0

∂

2

E

∂x∂t

.

Combining these two equations, we get

∂

2

B

∂t

2

=

1

ε

0

µ

0

∂

2

B

∂x

2

= c

2

∂

2

B

∂x

2

.