Proving differentiability of a function from condition of diff. of other function

Question

The following question says:

Let $$\phi(t) = \begin{cases} \dfrac{sin(t)}{t} & \text{if $t\neq 0 $} \\ 1 & \text{if $t=0$} \end{cases}$$

Show that $\phi$ is differentiable on $\mathbb R$.Let

$$f(x,y) = \begin{cases} \dfrac{cos\text{x}-cos\text{y}}{x-y} & \text{if $x\neq y $} \\ 0 & \text{ otherwise} \end{cases}$$

Express $f$ in terms of $\phi$ and show that $f$ is differentiable on $\mathbb R^2$.

I solved that $\phi$ is differentiable .

Now to express $f$ in terms of $\phi$ ,do we have to use formula for $cos\text{x}-cos\text{y}$ ....Also further does composition of differentiable functions is differentiable function?that we can use to show differentiability of $f$...

Please help....

Answer 1

Use trigonometric identities:

$$\cos x-\cos y=-2\sin\frac{x+y}2\,\sin\frac{x-y}2$$

Ans yes: composition of differentiable functions is differentiable, even in several variables.

Adding on request:

$$\frac{\cos x-\cos y}{x-y}=-\frac{\sin\frac{x-y}2}{\frac{x-y}2}\;\sin\frac{x+y}2\xrightarrow[(x,y)\to (0,0)]{}-1\cdot\sin 0=0$$