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

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 $f$...

$$\cos x-\cos y=-2\sin\frac{x+y}2\,\sin\frac{x-y}2$$
$$\frac{\cos x-\cos y}{x-y}=-\frac{\sin\frac{x-y}2}{\frac{x-y}2}\;\sin\frac{x+y}2\xrightarrow[(x,y)\to (0,0)]{}-1\cdot\sin 0=0$$
• thanks for hint...Any idea how to prove diff. of $f$.. – patang Oct 15 '14 at 14:55