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Bananainas
@Bananainas
July 2022
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Полное решение необязательно.
Найдите f'(x0), если f(x0)=(x^2+3x-4)^5-sinПx, x0=1
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sedinalana
Verified answer
F`(x)=5*(x^2+3x-4)^4*(2x+3)-πcosπx
x0=1
f`(1)=5*(1+3-4)^4*(2+3)-πcosπ=5*0*5-π*(-1)=π
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Verified answer
F`(x)=5*(x^2+3x-4)^4*(2x+3)-πcosπxx0=1
f`(1)=5*(1+3-4)^4*(2+3)-πcosπ=5*0*5-π*(-1)=π