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seb12
Korzystasz z reguły de l'Hospital'a
![f(x)=e^{(x-1)^2}-1\\ f'(x)=[e^{(x-1)^2}-1]'=[e^{(x-1)^2}]'-0=[e^{(x-1)^2}]'\\ u=(x-1)^2\\ \frac{d}{du}f(x)*\frac{du}{dx}=\frac{d}{dx}f(x)](https://tex.z-dn.net/?f=f%28x%29%3De%5E%7B%28x-1%29%5E2%7D-1%5C%5C%0Af%27%28x%29%3D%5Be%5E%7B%28x-1%29%5E2%7D-1%5D%27%3D%5Be%5E%7B%28x-1%29%5E2%7D%5D%27-0%3D%5Be%5E%7B%28x-1%29%5E2%7D%5D%27%5C%5C%0Au%3D%28x-1%29%5E2%5C%5C%0A%5Cfrac%7Bd%7D%7Bdu%7Df%28x%29%2A%5Cfrac%7Bdu%7D%7Bdx%7D%3D%5Cfrac%7Bd%7D%7Bdx%7Df%28x%29 "f(x)=e^{(x-1)^2}-1\\ f'(x)=[e^{(x-1)^2}-1]'=[e^{(x-1)^2}]'-0=[e^{(x-1)^2}]'\\ u=(x-1)^2\\ \frac{d}{du}f(x)*\frac{du}{dx}=\frac{d}{dx}f(x)")
![f'(x)=(e^u)'*[(x-1)^2]'=e^u*[(x-1)^2]'\\ v=x-1\\ f'(x)=e^{(x-1)^2}*(v^2)'*(x-1)'=e^{(x-1)^2}*2(x-1)*1\\ f'(x)=2(x-1)e^{(x-1)^2}](https://tex.z-dn.net/?f=f%27%28x%29%3D%28e%5Eu%29%27%2A%5B%28x-1%29%5E2%5D%27%3De%5Eu%2A%5B%28x-1%29%5E2%5D%27%5C%5C%0Av%3Dx-1%5C%5C%0Af%27%28x%29%3De%5E%7B%28x-1%29%5E2%7D%2A%28v%5E2%29%27%2A%28x-1%29%27%3De%5E%7B%28x-1%29%5E2%7D%2A2%28x-1%29%2A1%5C%5C%0Af%27%28x%29%3D2%28x-1%29e%5E%7B%28x-1%29%5E2%7D "f'(x)=(e^u)'*[(x-1)^2]'=e^u*[(x-1)^2]'\\ v=x-1\\ f'(x)=e^{(x-1)^2}*(v^2)'*(x-1)'=e^{(x-1)^2}*2(x-1)*1\\ f'(x)=2(x-1)e^{(x-1)^2}")
![g'(x)=[\sin^2(\pi x)]'=\\ u=\sin(\pi x)\\ =\frac{d}{du}u^2*\frac{d}{dx}[\sin(\pi x)]=2\sin(\pi x)*[\sin(\pi x)]'=\\ v=\pi x\\ =2\sin(\pi x)*\frac{d}{du}\sin u*\frac{d}{dx}(\pi x)=2\sin(\pi x)*\cos (\pi x)*\pi \\ =2\pi \sin(\pi x)\cos (\pi x)=\pi \sin(2\pi x)\\ Nadal\ wychodzi\ nam \ 0\ po podstawieniu\ 1,\ wiec\ rozniczkujemy\ dalej.\\ \\ Robisz\ to\ identycznie,\ bierzesz\ kolejna\ pochodna\ z\ g'(x)\\Zas\ pochodna\ z\ f'(x):\\ f''(x)=2e^{(x-1)^2}(2x^2-4x+3)\\ \\g''=[tex]\lim_{x\to 1}\frac{e^{(x-1)^2}-1}{\sin^2 (\pi x)}=\lim_{x\to 1}\frac{2e^{(x-1)^2}(2x^2-4x+3)}{2\pi ^2\cos(2\pi x)}=\\ \\=\frac{2e^{(1-1)^2}(2*1^2-4*1+3)}{2\pi ^2\cos(2\pi*1)}=\frac{2}{2\pi ^2}=\boxed{\underline{\frac{1}{\pi ^2}}}](https://tex.z-dn.net/?f=g%27%28x%29%3D%5B%5Csin%5E2%28%5Cpi+x%29%5D%27%3D%5C%5C%0Au%3D%5Csin%28%5Cpi+x%29%5C%5C%0A%3D%5Cfrac%7Bd%7D%7Bdu%7Du%5E2%2A%5Cfrac%7Bd%7D%7Bdx%7D%5B%5Csin%28%5Cpi+x%29%5D%3D2%5Csin%28%5Cpi+x%29%2A%5B%5Csin%28%5Cpi+x%29%5D%27%3D%5C%5C%0Av%3D%5Cpi+x%5C%5C%0A%3D2%5Csin%28%5Cpi+x%29%2A%5Cfrac%7Bd%7D%7Bdu%7D%5Csin+u%2A%5Cfrac%7Bd%7D%7Bdx%7D%28%5Cpi+x%29%3D2%5Csin%28%5Cpi+x%29%2A%5Ccos+%28%5Cpi+x%29%2A%5Cpi+%5C%5C%0A%3D2%5Cpi+%5Csin%28%5Cpi+x%29%5Ccos+%28%5Cpi+x%29%3D%5Cpi+%5Csin%282%5Cpi+x%29%5C%5C%0ANadal%5C+wychodzi%5C+nam+%5C+0%5C+po+podstawieniu%5C+1%2C%5C+wiec%5C+rozniczkujemy%5C+dalej.%5C%5C%0A%5C%5C%0ARobisz%5C+to%5C+identycznie%2C%5C+bierzesz%5C+kolejna%5C+pochodna%5C+z%5C+g%27%28x%29%5C%5CZas%5C+pochodna%5C+z%5C+f%27%28x%29%3A%5C%5C+f%27%27%28x%29%3D2e%5E%7B%28x-1%29%5E2%7D%282x%5E2-4x%2B3%29%5C%5C%C2%A0%5C%5Cg%27%27%3D%5Btex%5D%5Clim_%7Bx%5Cto+1%7D%5Cfrac%7Be%5E%7B%28x-1%29%5E2%7D-1%7D%7B%5Csin%5E2+%28%5Cpi+x%29%7D%3D%5Clim_%7Bx%5Cto+1%7D%5Cfrac%7B2e%5E%7B%28x-1%29%5E2%7D%282x%5E2-4x%2B3%29%7D%7B2%5Cpi+%5E2%5Ccos%282%5Cpi+x%29%7D%3D%5C%5C%0A%5C%5C%3D%5Cfrac%7B2e%5E%7B%281-1%29%5E2%7D%282%2A1%5E2-4%2A1%2B3%29%7D%7B2%5Cpi+%5E2%5Ccos%282%5Cpi%2A1%29%7D%3D%5Cfrac%7B2%7D%7B2%5Cpi+%5E2%7D%3D%5Cboxed%7B%5Cunderline%7B%5Cfrac%7B1%7D%7B%5Cpi+%5E2%7D%7D%7D "g'(x)=[\sin^2(\pi x)]'=\\ u=\sin(\pi x)\\ =\frac{d}{du}u^2*\frac{d}{dx}[\sin(\pi x)]=2\sin(\pi x)*[\sin(\pi x)]'=\\ v=\pi x\\ =2\sin(\pi x)*\frac{d}{du}\sin u*\frac{d}{dx}(\pi x)=2\sin(\pi x)*\cos (\pi x)*\pi \\ =2\pi \sin(\pi x)\cos (\pi x)=\pi \sin(2\pi x)\\ Nadal\ wychodzi\ nam \ 0\ po podstawieniu\ 1,\ wiec\ rozniczkujemy\ dalej.\\ \\ Robisz\ to\ identycznie,\ bierzesz\ kolejna\ pochodna\ z\ g'(x)\\Zas\ pochodna\ z\ f'(x):\\ f''(x)=2e^{(x-1)^2}(2x^2-4x+3)\\ \\g''=[tex]\lim_{x\to 1}\frac{e^{(x-1)^2}-1}{\sin^2 (\pi x)}=\lim_{x\to 1}\frac{2e^{(x-1)^2}(2x^2-4x+3)}{2\pi ^2\cos(2\pi x)}=\\ \\=\frac{2e^{(1-1)^2}(2*1^2-4*1+3)}{2\pi ^2\cos(2\pi*1)}=\frac{2}{2\pi ^2}=\boxed{\underline{\frac{1}{\pi ^2}}}")
\\ Zas\ pochodna\ z\ f'(x):\\ f''(x)=2e^{(x-1)^2}(2x^2-4x+3)\\[/tex]
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