Given\\ (f(x)=3x(x-1)^{5} compute \\ (f\`\`\`(x)\\)
f(x)=3x(x−1)5f(x)=3x(x-1)^5f(x)=3x(x−1)5
f′(x)=3(x−1)5+3x⋅5(x−1)4=f'(x)=3(x-1)^5+3x\cdot 5(x-1)^4=f′(x)=3(x−1)5+3x⋅5(x−1)4=
=3(x−1)4(x−1+5x)=3(x−1)4(6x−1)=3(x-1)^4(x-1+5x)=3(x-1)^4(6x-1)=3(x−1)4(x−1+5x)=3(x−1)4(6x−1)
f′′(x)=3⋅4(x−1)3(6x−1)+3(x−1)4⋅6=f''(x)=3\cdot 4(x-1)^3(6x-1)+3(x-1)^4\cdot 6=f′′(x)=3⋅4(x−1)3(6x−1)+3(x−1)4⋅6=
=6(x−1)3(12x−2+3x−3)=30(x−1)3(3x−1)=6(x-1)^3(12x-2+3x-3)=30(x-1)^3(3x-1)=6(x−1)3(12x−2+3x−3)=30(x−1)3(3x−1)
f′′′(x)=30⋅3(x−1)2(3x−1)+30(x−1)3⋅3=f'''(x)=30\cdot 3(x-1)^2(3x-1)+30(x-1)^3\cdot 3=f′′′(x)=30⋅3(x−1)2(3x−1)+30(x−1)3⋅3=
=90(x−1)2(3x−1+x−1)=180(x−1)2(2x−1)=90(x-1)^2(3x-1+x-1)=180(x-1)^2(2x-1)=90(x−1)2(3x−1+x−1)=180(x−1)2(2x−1)
Answer: f′′′(x)=180(x−1)2(2x−1)f'''(x)=180(x-1)^2(2x-1)f′′′(x)=180(x−1)2(2x−1)
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