## Task Condition

Based on the definition of the derivative, find $f^{\prime}(0)$:

$$
f(x)=\left\{\begin{array}{c}
\frac{2^{\operatorname{tg} x}-2^{\sin x}}{x^{2}}, x \neq 0 \\
0, x=0
\end{array}\right.
$$
## Solution

By definition, the derivative at the point $x=0$:

$f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}$

Based on the definition, we find:

$$
\begin{aligned}
& f^{\prime}(0)=\lim _{\Delta x \rightarrow 0} \frac{f(0+\Delta x)-f(0)}{\Delta x}=\lim _{\Delta x \rightarrow 0}\left(\frac{2^{\operatorname{tg} \Delta x}-2^{\sin \Delta x}}{\Delta x^{2}}-0\right) / \Delta x= \\
& =\lim _{\Delta x \rightarrow 0} \frac{2^{\operatorname{tg} \Delta x}-2^{\sin \Delta x}}{\Delta x^{3}}=\lim _{\Delta x \rightarrow 0} \frac{2^{\operatorname{tg} \Delta x}-2^{\sin \Delta x \cdot \cos \Delta x / \cos \Delta x}}{\Delta x^{3}}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{2^{\operatorname{tg} \Delta x}-2^{\operatorname{tg} \Delta x \cdot \cos \Delta x}}{\Delta x^{3}}=\lim _{\Delta x \rightarrow 0} \frac{-2^{\operatorname{tg} \Delta x}\left(2^{\operatorname{tg} \Delta x \cdot \cos \Delta x-\operatorname{tg} \Delta x}-1\right)}{\Delta x^{3}}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{-2^{\operatorname{tg} \Delta x}\left(2^{\operatorname{tg} \Delta x(1-\cos \Delta x)}-1\right)}{\Delta x^{3}}=\lim _{\Delta x \rightarrow 0} \frac{-2^{\operatorname{tg} \Delta x}\left(2^{-2 \operatorname{tg} \Delta x \cdot \sin ^{2}(\Delta x / 2)}-1\right)}{\Delta x^{3}}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{-2^{\operatorname{tg} \Delta x}\left(\left(e^{\ln 2}\right)^{-2 \operatorname{tg} \Delta x \cdot \sin ^{2}(\Delta x / 2)}-1\right)}{\Delta x^{3}}= \\
& =\lim _{\Delta x \rightarrow 0} \frac{-2^{\operatorname{tg} \Delta x}\left(e^{-2 \ln 2 \cdot \operatorname{tg} \Delta x \cdot \sin ^{2}(\Delta x / 2)}-1\right)}{\Delta x^{3}}=
\end{aligned}
$$

Using the substitution of equivalent infinitesimals:

$e^{-2 \ln 2 \cdot \operatorname{tg} \Delta x \cdot \sin ^{2} \frac{\Delta x}{2}}-1 \sim-2 \ln 2 \cdot \operatorname{tg} \Delta x \cdot \sin ^{2}(\Delta x / 2)$, as

$\Delta x \rightarrow 0\left(-2 \ln 2 \cdot \operatorname{tg} \Delta x \cdot \sin ^{2} \frac{\Delta x}{2} \rightarrow 0\right)$

We get:

$=\lim _{\Delta x \rightarrow 0} \frac{-2^{\operatorname{tg} \Delta x}\left(-2 \ln 2 \cdot \operatorname{tg} \Delta x \cdot \sin ^{2} \frac{\Delta x}{2}\right)}{\Delta x^{3}}=$

Using the substitution of equivalent infinitesimals:

$\operatorname{tg} \Delta x \sim \Delta x$, as $\Delta x \rightarrow 0$

$\sin \frac{\Delta x}{2} \sim \frac{\Delta x}{2}$, as $\Delta x \rightarrow 0\left(\frac{\Delta x}{2} \rightarrow 0\right)$

We get:

$$
\begin{aligned}
& =\lim _{\Delta x \rightarrow 0} \frac{-2^{\operatorname{tg} \Delta x}\left(-2 \ln 2 \cdot \Delta x \cdot\left(\frac{\Delta x}{2}\right)^{2}\right)}{\Delta x^{3}}=\lim _{\Delta x \rightarrow 0} \frac{2^{\operatorname{tg} \Delta x} \cdot 2 \ln 2 \cdot \Delta x^{3} \cdot \frac{1}{4}}{\Delta x^{3}}= \\
& =\lim _{\Delta x \rightarrow 0} 2^{\operatorname{tg} \Delta x} \cdot \ln 2 \cdot \frac{1}{2}=2^{\operatorname{tg} 0} \cdot \ln 2 \cdot \frac{1}{2}= \\
& =2^{0} \cdot \ln \sqrt{2}=\ln \sqrt{2} \\
& \text { Therefore, } f^{\prime}(0)=\ln \sqrt{2}
\end{aligned}
$$

## Problem Kuznetsov Differentiation $2-24$