Найти f `(x0) , если f(x) = 1/2 sin2x, x0= п/8
Найти f `(x0) , если f(x) = (3x-5)^3 + 1/ (3-x)^2, x0=2
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1)
![x_0= frac{ pi }{8} x_0= frac{ pi }{8}](https://tex.z-dn.net/?f=x_0%3D+frac%7B+pi+%7D%7B8%7D+)
![f'(x) =( frac{1}{2} sin2x)'= frac{1}{2} cos2x*(2x)'= frac{1}{2} cos2x*2=cos2x f'(x) =( frac{1}{2} sin2x)'= frac{1}{2} cos2x*(2x)'= frac{1}{2} cos2x*2=cos2x](https://tex.z-dn.net/?f=f%27%28x%29+%3D%28++frac%7B1%7D%7B2%7D++sin2x%29%27%3D+frac%7B1%7D%7B2%7D++cos2x%2A%282x%29%27%3D+frac%7B1%7D%7B2%7D++cos2x%2A2%3Dcos2x)
![f'( frac{ pi }{8} )=cos(2* frac{ pi }{8}) =cos frac{ pi }{4} = frac{ sqrt{2} }{2} f'( frac{ pi }{8} )=cos(2* frac{ pi }{8}) =cos frac{ pi }{4} = frac{ sqrt{2} }{2}](https://tex.z-dn.net/?f=f%27%28+frac%7B+pi+%7D%7B8%7D+%29%3Dcos%282%2A+frac%7B+pi+%7D%7B8%7D%29+%3Dcos+frac%7B+pi+%7D%7B4%7D+%3D+frac%7B+sqrt%7B2%7D+%7D%7B2%7D+)
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![x_0=2 x_0=2](https://tex.z-dn.net/?f=x_0%3D2)
![f'(x) =( (3x-5)^3 + frac{1}{(3-x)^2} )'=((3x-5)^3)' + ( frac{1}{(3-x)^2} )'= f'(x) =( (3x-5)^3 + frac{1}{(3-x)^2} )'=((3x-5)^3)' + ( frac{1}{(3-x)^2} )'=](https://tex.z-dn.net/?f=f%27%28x%29+%3D%28+%283x-5%29%5E3+%2B++frac%7B1%7D%7B%283-x%29%5E2%7D+%29%27%3D%28%283x-5%29%5E3%29%27+%2B+%28+frac%7B1%7D%7B%283-x%29%5E2%7D+%29%27%3D)
![=3*(3x-5)^2*3+ frac{0-2(3-x)*(-1)}{(3-x)^4} =9*(3x-5)^2+ frac{2(3-x)}{(3-x)^4} = =3*(3x-5)^2*3+ frac{0-2(3-x)*(-1)}{(3-x)^4} =9*(3x-5)^2+ frac{2(3-x)}{(3-x)^4} =](https://tex.z-dn.net/?f=%3D3%2A%283x-5%29%5E2%2A3%2B+frac%7B0-2%283-x%29%2A%28-1%29%7D%7B%283-x%29%5E4%7D+%3D9%2A%283x-5%29%5E2%2B+frac%7B2%283-x%29%7D%7B%283-x%29%5E4%7D+%3D)
![=9*(3x-5)^2+ frac{2}{(3-x)^3} =9*(3x-5)^2+ frac{2}{(3-x)^3}](https://tex.z-dn.net/?f=%3D9%2A%283x-5%29%5E2%2B+frac%7B2%7D%7B%283-x%29%5E3%7D)
![f'(2)=9*(3*2-5)^2+ frac{2}{(3-2)^3}=9+2=11 f'(2)=9*(3*2-5)^2+ frac{2}{(3-2)^3}=9+2=11](https://tex.z-dn.net/?f=f%27%282%29%3D9%2A%283%2A2-5%29%5E2%2B+frac%7B2%7D%7B%283-2%29%5E3%7D%3D9%2B2%3D11)
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