Помогите пожалуйста !!! Найдите производные функции:
а) f (x) = (4-3x)*100
б) y(x) = корень x^4+1
в) f (x) = sin2x-cos3x
г) y (x) = sin^2x
Ответы
Ответ дал:
7
Немного теории:
![(x^n)'=nx^{n-1}\\\sqrt{x}=x^\frac{1}{2}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\\(sinx)'=cosx\\(cosx)'=-sinx\\(f(g(x))'=f'(g)*g'(x) (x^n)'=nx^{n-1}\\\sqrt{x}=x^\frac{1}{2}=\frac{1}{2}x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\\(sinx)'=cosx\\(cosx)'=-sinx\\(f(g(x))'=f'(g)*g'(x)](https://tex.z-dn.net/?f=%28x%5En%29%27%3Dnx%5E%7Bn-1%7D%5C%5C%5Csqrt%7Bx%7D%3Dx%5E%5Cfrac%7B1%7D%7B2%7D%3D%5Cfrac%7B1%7D%7B2%7Dx%5E%7B%5Cfrac%7B1%7D%7B2%7D-1%7D%3D%5Cfrac%7B1%7D%7B2%7Dx%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%3D%5Cfrac%7B1%7D%7B2%5Csqrt%7Bx%7D%7D%5C%5C%28sinx%29%27%3Dcosx%5C%5C%28cosx%29%27%3D-sinx%5C%5C%28f%28g%28x%29%29%27%3Df%27%28g%29%2Ag%27%28x%29)
![f'(x)=((4-3x)*100)'=100*(4-3x)'=100*(0-3)=-300\\\\y'(x)=(\sqrt{x^4+1})'=\frac{1}{2\sqrt{x^4+1}}*(x^4+1)'=\frac{1}{2\sqrt{x^4+1}}*4x^3=\frac{2x^3}{\sqrt{x^4+1}}\\\\f'(x)=(sin2x-cos3x)'=(sin2x)'-(cos3x)'=\\=cos2x*(2x)'-(-sin3x)*(3x)'=2cos2x+3sin3x\\\\y'(x)=((sinx)^2)'=2sinx*(sinx)'=2sinx*cosx=sin2x f'(x)=((4-3x)*100)'=100*(4-3x)'=100*(0-3)=-300\\\\y'(x)=(\sqrt{x^4+1})'=\frac{1}{2\sqrt{x^4+1}}*(x^4+1)'=\frac{1}{2\sqrt{x^4+1}}*4x^3=\frac{2x^3}{\sqrt{x^4+1}}\\\\f'(x)=(sin2x-cos3x)'=(sin2x)'-(cos3x)'=\\=cos2x*(2x)'-(-sin3x)*(3x)'=2cos2x+3sin3x\\\\y'(x)=((sinx)^2)'=2sinx*(sinx)'=2sinx*cosx=sin2x](https://tex.z-dn.net/?f=f%27%28x%29%3D%28%284-3x%29%2A100%29%27%3D100%2A%284-3x%29%27%3D100%2A%280-3%29%3D-300%5C%5C%5C%5Cy%27%28x%29%3D%28%5Csqrt%7Bx%5E4%2B1%7D%29%27%3D%5Cfrac%7B1%7D%7B2%5Csqrt%7Bx%5E4%2B1%7D%7D%2A%28x%5E4%2B1%29%27%3D%5Cfrac%7B1%7D%7B2%5Csqrt%7Bx%5E4%2B1%7D%7D%2A4x%5E3%3D%5Cfrac%7B2x%5E3%7D%7B%5Csqrt%7Bx%5E4%2B1%7D%7D%5C%5C%5C%5Cf%27%28x%29%3D%28sin2x-cos3x%29%27%3D%28sin2x%29%27-%28cos3x%29%27%3D%5C%5C%3Dcos2x%2A%282x%29%27-%28-sin3x%29%2A%283x%29%27%3D2cos2x%2B3sin3x%5C%5C%5C%5Cy%27%28x%29%3D%28%28sinx%29%5E2%29%27%3D2sinx%2A%28sinx%29%27%3D2sinx%2Acosx%3Dsin2x)
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