Hipi Zhdripi i Matematikës/1282

Stampa:Hipi Zhdripi i Matematikës kryefleta Stampa:V ë r t e t i m Forma implicite e këtij funksioni është $x=\mathrm{t}\mathrm{g}y$ i cili kur derivohet marrim:

$1=\frac{y^{\prime }}{{\cos }^{2}y}{y}^{\prime }={\cos }^{2}y=\frac{1}{1+{\mathrm{t}\mathrm{g}}^{2}y}$.

Kur në këtë barazi e zëvendësojmë $\mathrm{t}\mathrm{g}y=x$ del:

$y^{\prime }=\frac{1}{1+x^2}\mathrm{o}\mathrm{s}\mathrm{e}(\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{g}x{)}^{\prime }=\frac{1}{1+x^2}$.

Stampa:Dygishta Për funksionin e përbërë $y=\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{g}u$ kemi:

$(\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{g}u{)}^{\prime }=\frac{u^{\prime }}{1+u^2}$ . (59a)

Stampa:T e o r e m a , pra

Stampa:V ë r t e t i m Marrim identitetin $\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{g}x+\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{t}\mathrm{g}x=\frac{\pi }{2}$. Me derivimin e tij përftohet: $(\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{g}x{)}^{\prime }+(\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{t}\mathrm{g}x{)}^{\prime }=0$ prej nga del:

$(\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{t}\mathrm{g}x{)}^{\prime }=-(\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{g}x{)}^{\prime }=-\frac{1}{1+x^2}$.

Stampa:Dygishta Për funksionin e përbërë $y=\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{t}\mathrm{g}u$ kemi formulën:

$(\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{t}\mathrm{g}u{)}^{\prime }=-\frac{1}{1+u^2}$ . (60a)

Stampa:S h e m b u l l i Derivati i funksionit $y=\mathrm{a}\mathrm{r}\mathrm{c}\sin \sqrt{\sin x}$ është:

$y^{\prime }=(\mathrm{a}\mathrm{r}\mathrm{c}\sin \sqrt{\sin x}{)}^{\prime }=\frac{(\sqrt{\sin x}{)}^{\prime }}{\sqrt{1-\sin x}}=\frac{\cos x}{2\sqrt{\sin x-{\sin }^{2}x}}.$

Stampa:S h e m b u l l i Derivati i funksionit $y=\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{g}\frac{4\sin x}{3+5\cos x}$ është:

$\begin{array}{cc}{y}^{\prime }& =\left(\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{t}\mathrm{g}\frac{4\sin x}{3+5\cos x}\right)=\frac{\left(\frac{4\sin x}{3+5\cos x}\right)}{1+{\left(\frac{4\sin x}{3+5\cos x}\right)}^2}=\cdots =\frac{4(5+3\cos x)}{(5+3\cos x{)}^2}\\ & =\frac{4}{5+3\cos x}.\end{array}$