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Penguin200
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Pe: 2019-01-08T09:20:01+02:00 In: In:

Derivata unei functii intr-un punct

Buna ziua!As avea nevoie de ajutor la urmatorul exercitiu:
Se da functia f(x)=\sqrt[5]{x^3-tg^3x}.Se cere derivata in punctul 0.
Multumesc.

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  1. FaN.Anduu
    user (0)

    Foloseste definitia derivatei.
    f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}
    \lim_{x\rightarrow 0}\frac{\sqrt[5]{x^{3}-tg^{3}(x)}}{x}=\lim_{x\rightarrow 0}\sqrt[5]{\frac{x^{3}-tg^{3}(x)}{x^{5}}}=\sqrt[5]{\lim_{x\rightarrow 0}\frac{x^{3}-tg^{3}(x)}{x^{5}}}
    Ce e sub radical notezi cu L deci vei avea \sqrt[5]{L}
    L=\lim_{x\rightarrow 0}\frac{(x-tg(x))\cdot (x^{2}+xtg(x)+tg^{2}(x))}{x^{5}}
    L=\lim_{x\rightarrow 0}\frac{x-tg(x)}{x^{3}}\cdot (\frac{x^{2}}{x^{2}}+\frac{x}{x}+\frac{tg(x)}{x}+\frac{tg^{2}(x)}{x^{2}})
    L=3\cdot \lim_{x\rightarrow 0}\frac{x-tg(x)}{x^{3}}=3\cdot \lim_{x\rightarrow 0}\frac{1-\frac{1}{cos^{2}(x)}}{3x^{2}}=\lim_{x\rightarrow 0}\frac{\frac{cos^{2}(x)-1}{cos^{2}(x)}}{x^{2}}
    L=\lim_{x\rightarrow 0}\frac{1}{cos^{2}(x)}\cdot \lim_{x\rightarrow 0}\frac{cos^{2}(x)-1}{x^{2}}
    Prima limita tinde la 1.A doua limita e -1 dupa ce aplici L’Hopital de 2 ori.Deci L=1*(-1)=-1
    Deci limita initiala e \sqrt[5]{L}=\sqrt[5]{-1}=-1

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