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Extend the displayed function by $f(0)=0$. Its difference quotient at zero is $h\sin (1/h)$, which tends to zero. Away from zero its derivative is $2x\sin (1/x)-\cos (1/x)$. At $x_k=1/(2\pi k)$ this derivative equals $-1$, whereas at $y_k=1/((2k+1)\pi )$ it equals $+1$. Thus differentiability implies continuity of the function but does not imply continuity of its derivative.

A real function is differentiable at an interior point $x$ when the finite limitexists, with $h$ restricted so $x+h$ lies in its domain. In quantified form, for every $\epsilon >0$ there is $\delta >0$ such that $0<|h|<\delta$ implies that the difference quotient differs from $f^{\prime }(x)$ by less than $\epsilon$.

To prove differentiability implies continuity, the convergent difference quotient is bounded by $|f^{\prime }(x)|+1$ for sufficiently small nonzero $h$. Thus $|f(x+h)-f(x)|\le (|f^{\prime }(x)|+1)|h|\to 0$. This is exactly continuity at $x$.

At zero, the particular function has difference quotient $h\sin (1/h)$. The bound $|h\sin (1/h)|\le |h|$ and the squeeze theorem giveFor $x\ne 0$, the product rule and chain rule giveAt $x_k=1/(2\pi k)$ this equals $-1$, although $x_k\to 0$ and $f^{\prime }(0)=0$. At $y_k=1/((2k+1)\pi )$ it equals $+1$. Hence the derivative has no limit at zero and is not continuous there. This demonstrates that differentiability does not imply continuity of the derivative, even though the function itself is continuous.

True. Differentiability implies continuity.

On the nonzero $x$ axis the function equals one, whereas its assigned value at the origin is zero. Therefore $f(x,0)$ does not tend to $f(0,0)$ as $x\to 0$. It is not a continuous function there, and differentiability implies continuity shows

A function is a differentiable function at $a_0=(a,b)$ if there is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ withThe map $L=D{f}_{a_0}$ is its derivative; equivalently $L(h_1,h_2)=A{h}_1+B{h}_2$ for fixed constants. Since every such linear map is bounded, $|f(a_0+h)-f(a_0)|\le \Vert L\Vert \Vert h\Vert +|r(h)|\to 0$. This proves differentiability implies continuity, not merely continuity along each coordinate axis.

By differentiability implies continuity, $f(x+h)\to f(x)$. Add and subtract $f(x+h)g(x)$ in the difference quotient:The two quotient limits exist by the given differentiability. Taking the limit and using continuity yields the product rule:Thus the product is differentiable at the specified point; differentiability elsewhere is not needed.

At zero, $|f(x)-f(0)|=|x|\to 0$, so the function is continuous. At a nonzero $x_0$, approach along rational and irrational sequences; density of the rational numbers and density of the irrational numbers give limiting values $x_0$ and $-x_0$, which are different. Thus the continuity set is $\boxed{\{0\}}$.

At zero, the difference quotient is one for rational nonzero increments and minus one for irrational increments, so the derivative does not exist. Elsewhere differentiability implies continuity, and continuity already fails. Thus the differentiability set is $\boxed{\varnothing }$.