We present a definition of local fractional derivative using kernels.

**Definition 2.1:** *Let k* : [*a, b*] → ℝ *be a* *continuous nonnegative map such that k(t) ≠* 0, *whenever* *t > a. Given a function* ƒ : [*a, b*] → ℝ *and α* ∈ (0, 1) a real, we say that ƒ is *α-differentiable at t > a, with respect to kernel k, if the limit*
$${f}^{\left(\alpha \right)}\left(t\right):=\underset{\u03f5\to 0}{lim}\frac{f\left(t+\u03f5k{\left(t\right)}^{1-\alpha}\right)-f\left(t\right)}{\u03f5}$$(1)*exists. The α-derivative at t = a is defined by*
$${f}^{\left(\alpha \right)}\left(a\right):=\underset{t\to {a}^{+}}{lim}{f}^{\left(\alpha \right)}\left(t\right),$$*if the limit exists*.Consider the limit *α →* 1^{–}. In this case, for *t > a*, we obtain the classical definition for derivative of a function, *f*^{(α)}(*t*) = ƒ′(*t*). Our definition is a more general concept, compared to others that we find in the literature. For example, taking *k*(*t*)* = t* and *a =* 0, we get the definition from [7, 8, 10, 11, 13] (also called conformable fractional derivative); when *k*(*t*)* = t — a*, the one from [1, 2, 18]; for *k*(*t*)* = t +* 1/Γ(*α*), the definition in [4, 5].The following result is trivial, and we omit the proof.

**Theorem 2.2:** *Let* ƒ : [*a, b*] → ℝ *be a differentiablefunction and t > a. Then*, ƒ *is a-differentiable at t and*
$${f}^{\left(\alpha \right)}\left(t\right)=k{\left(t\right)}^{1-\alpha}{f}^{\prime}\left(t\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t>a.$$*Also, if f′ is continuous at t = a, then*
$${f}^{\left(\alpha \right)}\left(a\right)=k{\left(a\right)}^{1-\alpha}{f}^{\prime}\left(a\right).$$However, there exist *α*-differentiable functions which are not differentiable in the usual sense. For example, consider the function $f\left(t\right)=\sqrt{t},$ with *t* ≥ 0. If we take the kernel *k*(*t*)* = t*, then ƒ^{(α)}(*t*)* =* l/2 *t*^{1/2-α}. Thus, for *α ∈* (0,1/2), ƒ^{(α)}(0) = 0 and for *α =* 1/2, ƒ^{(α)}(0) = 1/2. In general, if we consider the function $f\left(t\right)=\sqrt[n]{t},$ with *t* ≥ 0 and *n* ∈ ℕ \ {1}, we have *f*^{(a)}(*t*)* =* 1*/nt* ^{1/n-α} and so *f*^{(α)}(0) = 0 if *α* ∈ (0, 1*/n*) and for *α =* 1*/n*, ƒ^{(α)}(0) = 1*/n*.

**Theorem 2.3:** *If f*^{(α)}(*t*)* exists for t > a, then* ƒ *is differentiable at t and*
$${f}^{\prime}\left(t\right)=k{\left(t\right)}^{\alpha -1}{f}^{\left(\alpha \right)}\left(t\right).$$

**Proof:** *It follows from
$${f}^{\prime}\left(t\right)=\underset{\delta \to 0}{lim}\frac{f\left(t+\delta \right)-f\left(t\right)}{\delta}=k{\left(t\right)}^{\alpha -1}\underset{\u03f5\to 0}{lim}\frac{f\left(t+\u03f5k{\left(t\right)}^{1-\alpha}\right)-f\left(t\right)}{\u03f5}=k{\left(t\right)}^{\alpha -1}{f}^{\left(\alpha \right)}\left(t\right).$$Of course we can not conclude anything at the initial point **t = a*, as was discussed before.Combining the two previous results, we have the main result of our paper.

**Theorem 2.4:** *A function* ƒ : [*a,b*] → ℝ *is α-differentiable at t > a if and only if it is differentiable at t. In that case, we have the relation*
$${f}^{\left(\alpha \right)}\left(t\right)=k{\left(t\right)}^{1-\alpha}{f}^{\prime}\left(t\right),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t>a.$$

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