Hipi Zhdripi i Matematikës/1280

Derivati i funksionit të përbërë $y=\cos u$ përcaktohet me formulën: $(\cos u{)}^{\prime }=-\sin u\cdot u^{\prime }$. (54a)        T e o r e m a  3.6.6. - Derivati i funksionit trigonometrik $y=tgx$ është i barabartë me $\frac{1}{{\cos }^{2}x}$, pra: $(\mathrm{t}\mathrm{g}x{)}^{\prime }=\frac{1}{{\cos }^{2}x}$. (55)        V ë r t e t i m Duke aplikuar formulat (42), (53), (54) marrim: $\begin{array}{cc}(\mathrm{t}\mathrm{g}x{)}^{\prime }& ={\left(\frac{\sin x}{\cos x}\right)}^{\prime }=\frac{(\sin x{)}^{\prime }\cos x-\sin x(\cos x{)}^{\prime }}{co{s}^{2}x}\\ & =\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}\end{array}$        T e o r e m a  3.6.7. - Derivati i funksionit trigonometrik $y=\mathrm{c}\mathrm{t}\mathrm{g}x$ është i barabartë me $-\frac{1}{si{n}^{2}x}$, pra: $(\mathrm{c}\mathrm{t}\mathrm{g}x{)}^{\prime }=-\frac{1}{{\sin }^{2}x}$. (56)        V ë r t e t i m Pasi që $\mathrm{c}\mathrm{t}\mathrm{g}x=\mathrm{t}\mathrm{g}(\frac{\pi }{2}-x)$, andaj del $(\mathrm{c}\mathrm{t}\mathrm{g}x{)}^{\prime }={\left[\mathrm{t}\mathrm{g}(\frac{\pi }{2}-x)\right]}^{\prime }=\frac{{(\frac{\pi }{2}-x)}^{\prime }}{{\cos }^2(\frac{\pi }{2}-x)}=-\frac{1}{{\sin }^{2}x}$         Për funksionin e përbërë $y=\mathrm{t}\mathrm{g}u$ kemi: $(\mathrm{t}\mathrm{g}u{)}^{\prime }=\frac{u^{\prime }}{{\cos }^{2}u}$. (55a)         Për funksionin e përbërë $y=\mathrm{c}\mathrm{t}\mathrm{g}u$ kemi $(\mathrm{c}\mathrm{t}\mathrm{g}u{)}^{\prime }=-\frac{u^{\prime }}{{\sin }^{2}u}$. (56a)        S h e m b u l l i  40. -  Derivati i funksionit $y=e^{\cos x}\sin x$ është:         $\begin{array}{cc}{y}^{\prime }& =(e^{\cos x}\sin x{)}^{\prime }=(e^{\cos x}{)}^{\prime }\sin x+e^{\cos x}(sinx{)}^{\prime }\\ & =e^{\cos x}(\cos x{)}^{\prime }\sin x+e^{\cos x}\cos x=e^{\cos x}(\cos x-{\sin }^{2}x).\end{array}$

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