Estimados expertos, su apoyo con esta duda de derivadas de orden superior

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Hola angel lara!

$$\begin{align}&f'(0)= \lim_{h\to 0} \frac{f(0+h)-f(0)}{h}=\lim_{h\to 0} \frac{\frac{\sin(0+h)}{h}-1}{h}=\\&\\&\lim_{h\to 0} \frac{\frac{\sin h}h -1}{h}= \frac{1-1} 0= \frac 0 0= \lim_{h\to 0}\frac{sinh -h}{h^2}= \frac  0 0=L'Hopital=\\&\\&\lim_{h\to 0} \frac{cosh-1}{2h}=\frac{1-1} 0= \frac 0 0=L'Hopital= \lim_{h\to 0} \frac{-sinh}{2}=  \frac 0 2 =0\end{align}$$

$$\begin{align}&f'(x)=\frac{xcosx-sinx}{x^2}\\&\\&f''(0)\\&\text{aplicando la definicion de derivada}\\&\\&f''(0) =\lim_{h \to 0} \frac{f'(0+h)-f'(0)}{h}= \lim_{h \to 0} \frac{\frac{hcosh-sinh}{h^2}-0}{h}=\\&\\&\lim_{h \to 0} \frac{hcosh-sinh}{h^3}=\frac 0 0=L`Hopital= \\&\\&\lim_{h \to 0} \frac{cosh+h(-sinh)-cosh}{3h^2}=\lim_{h \to 0}\frac{-hsinh}{3h^2}=\\&\\&\lim_{h \to 0}\frac{-sinh}{3h}= \frac 0 0= L'Hopital= \lim_{h \to 0} \frac{-cosh}{3}= - \frac 1 3\\&\end{align}$$

Saludos  y recuerda votar

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