Question Number 194338 by mathlove last updated on 04/Jul/23
Answered by Peace last updated on 04/Jul/23
![e^(cos(t)ln(2+sin(t))) −2=f(t) f in [0,x];∃c∈]0,x[ ⇒f(x)−f(0)=f′(c)(x−0) ⇒f′(c)=(((2+sin(x))^(cos(x)) −2)/x) f′(c)={−sin(c)ln(2+sin(c))+((cos^2 (c))/(2+sin(c)))}(2+sin(c))^(cos(c)) lim_(c→0) f′(c)=1](Q194340.png)
$${e}^{{cos}\left({t}\right){ln}\left(\mathrm{2}+{sin}\left({t}\right)\right)} −\mathrm{2}={f}\left({t}\right) \\ $$$$\left.{f}\:{in}\:\left[\mathrm{0},{x}\right];\exists{c}\in\right]\mathrm{0},{x}\left[\right. \\ $$$$\Rightarrow{f}\left({x}\right)−{f}\left(\mathrm{0}\right)={f}'\left({c}\right)\left({x}−\mathrm{0}\right) \\ $$$$\Rightarrow{f}'\left({c}\right)=\frac{\left(\mathrm{2}+{sin}\left({x}\right)\right)^{{cos}\left({x}\right)} −\mathrm{2}}{{x}} \\ $$$${f}'\left({c}\right)=\left\{−{sin}\left({c}\right){ln}\left(\mathrm{2}+{sin}\left({c}\right)\right)+\frac{{cos}^{\mathrm{2}} \left({c}\right)}{\mathrm{2}+{sin}\left({c}\right)}\right\}\left(\mathrm{2}+{sin}\left({c}\right)\right)^{{cos}\left({c}\right)} \\ $$$$\underset{{c}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}'\left({c}\right)=\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Answered by cortano12 last updated on 04/Jul/23
$$\:\:\: \\ $$$$ \:\mathrm{L}=\mathrm{2}.\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{sin}\:\mathrm{x}}{\mathrm{2}}\right)^{\mathrm{cos}\:\mathrm{x}} −\mathrm{1}}{\mathrm{x}} \\ $$$$\:\:\mathrm{L}\:=\mathrm{2}.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\left(\mathrm{1}+\frac{\mathrm{cos}\:\mathrm{x}\:\mathrm{sin}\:\mathrm{x}}{\mathrm{2}}\right)−\mathrm{1}}{\mathrm{x}} \\ $$$$\:\mathrm{L}=\:\mathrm{2}.\:\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{4x}}\right)=\:\begin{array}{|c|}{\mathrm{1}}\\\hline\end{array} \\ $$