Levi i desni izvod

Levi i desni izvod 08.12.2011. u 23:08 · pre 181 meseci #1

Kako da izracunam levi i desni izvod od $f(x)=\sqrt{sinx^{2}}$ ?

Re: Levi i desni izvod 08.12.2011. u 23:20 · pre 181 meseci #2

Desni

$f'\left(x^+\right)=\lim_{h\to 0+} \, \frac{\sqrt{\sin (x+h)^2}-\sqrt{\sin \left(x^2\right)}}{h}.$

Levi

$f'\left(x^-\right)=\lim_{h\to 0-} \, \frac{\sqrt{\sin (x+h)^2}-\sqrt{\sin \left(x^2\right)}}{h}.$

Jel tebe muči kako ovo da izračunaš?

Potpis

A mathematician is a blind man in a dark room looking for a black cat which isn't there.

Re: Levi i desni izvod 08.12.2011. u 23:24 · pre 181 meseci #3

Upravo tako :)

Re: Levi i desni izvod 08.12.2011. u 23:56 · pre 181 meseci #4

$f'\left(x^+\right)=\lim_{h\to 0+} \, \frac{\sqrt{\sin (x+h)^2}-\sqrt{\sin \left(x^2\right)}}{h}\left(\frac{\sqrt{\sin (x+h)^2}+\sqrt{\sin \left(x^2\right)}}{\sqrt{\sin (x+h)^2}+\sqrt{\sin \left(x^2\right)}}\right)=\lim_{h\to 0+} \, \frac{\sin (x+h)^2-\sin \left(x^2\right)}{h\left(\sqrt{\sin (x+h)^2}+\sqrt{\sin \left(x^2\right)}\right)}=\lim_{h\to 0+} \, \frac{2 \cos \left(x^2+h x+\frac{h^2}{2}\right)\sin \left(\frac{h^2}{2}+h x\right)}{h\left(\sqrt{\sin (x+h)^2}+\sqrt{\sin \left(x^2\right)}\right)}=$

$=\lim_{h\to 0+} \, \frac{2\cos \left(x^2+h x+\frac{h^2}{2}\right)}{\left(\sqrt{\sin (x+h)^2}+\sqrt{\sin \left(x^2\right)}\right)}\lim_{h\to 0+} \, \frac{\sin \left(\frac{h^2}{2}+h x\right)}{h}=\, \frac{2\cos \left(x^2\right)}{2\sqrt{\sin x^2}}\lim_{h\to 0+} \, \frac{\sin \left(\frac{h^2}{2}+h x\right)}{h}=\frac{2\cos \left(x^2\right)}{2\sqrt{\sin x^2}}\lim_{h\to 0+} \frac{\frac{h^2}{2}+h x+o\left(\left(\frac{h^2}{2}+h x\right)^2\right)}{h}=\frac{\cos x^2 }{\sqrt{\sin x^2}}x=\frac{x \cos x^2}{\sqrt{\sin x^2}}$

Na kraju sam ovo koristio $\sin (x)\sim x, x\to 0$ , valjda ti je to dozvoljeno.
A na početku $\sin (\alpha )-\sin (\beta )=2 \sin \left(\frac{\alpha }{2}-\frac{\beta }{2}\right) \cos \left(\frac{\alpha }{2}+\frac{\beta }{2}\right)$
Levi analogno.

_{[Ovu poruku je menjao SrdjanR271 dana 09.12.2011. u 02:06 GMT+1]}

Potpis

A mathematician is a blind man in a dark room looking for a black cat which isn't there.

Re: Levi i desni izvod 09.12.2011. u 02:14 · pre 181 meseci #5

Moglo je lakše, sad se setih.

$\frac{\cos x^2}{\sqrt{ \sin x^2}} \lim_{h\to 0+} \, \frac{\sin \left(\frac{h^2}{2}+h x\right)}{h}=\frac{\cos x^2}{\sqrt{ \sin x^2}} \lim_{h\to 0+} \, \frac{\sin \left(\frac{h^2}{2}+h x\right)}{\frac{h^2}{2}+h x} \frac{\left(\frac{h^2}{2}+h x\right)}{h}=\frac{\cos x^2}{\sqrt{ \sin x^2}} \lim_{h\to 0+} \, \frac{\sin \left(\frac{h^2}{2}+h x\right)}{\frac{h^2}{2}+h x}\lim_{h\to 0+} \frac{\left(\frac{h^2}{2}+h x\right)}{h}=\frac{\cos x^2}{\sqrt{ \sin x^2}}*1*\lim_{h\to 0+} \left(\frac{h}{2}+x\right)=\frac{x \cos x^2}{\sqrt{ \sin x^2}}$

Potpis

A mathematician is a blind man in a dark room looking for a black cat which isn't there.

Srodne teme