Toán hàm số y= (x ²+1) ³ có đạo hàm cấp 3 là ? 23/10/2021 By Adalyn hàm số y= (x ²+1) ³ có đạo hàm cấp 3 là ?
y’=3.(x²+1)².(x²+1)’=3.(x²+1)².2x=6x.(x²+1)²y”=(6x)'(x²+1)²+6x.((x²+1)²)’=6.(x²+1)²+6x.(2(x²+1).(x²+1)’)=6.(x²+1)²+6x.(2(x²+1).2x)=6.(x²+1)²+24x²(x²+1)y”’=(6(x²+1)²)’+(24x²(x²+1))’=6((x²+1)²)’+(24x²)'(x²+1)+24x²(x²+1)’=6(2(x²+1)(x²+1)’)+48x(x²+1)+24x².2x=6(2(x²+1)2x)+48x(x²+1)+48x³=24x(x²+1)+48x(x²+1)+48x³=72x(x²+1)+48x³ Trả lời
Đáp án: \(y”’ = 120{x^3} + 72x\) Giải thích các bước giải: \(\begin{array}{l}y = {\left( {{x^2} + 1} \right)^3}\\y’ = 3.2x.{\left( {{x^2} + 1} \right)^2} = 6x{\left( {{x^2} + 1} \right)^2}\\y” = 6{\left( {{x^2} + 1} \right)^2} + 2.2x\left( {{x^2} + 1} \right).6x\\ = 6{\left( {{x^2} + 1} \right)^2} + 24{x^2}\left( {{x^2} + 1} \right)\\ = \left( {{x^2} + 1} \right)\left( {6{x^2} + 6 + 24{x^2}} \right)\\ = \left( {{x^2} + 1} \right)\left( {30{x^2} + 6} \right)\\y”’ = 2x\left( {30{x^2} + 6} \right) + 30.2.x\left( {{x^2} + 1} \right)\\ = 60{x^3} + 12x + 60{x^3} + 60x\\ = 120{x^3} + 72x\end{array}\) Trả lời
y’=3.(x²+1)².(x²+1)’=3.(x²+1)².2x=6x.(x²+1)²y”=(6x)'(x²+1)²+6x.((x²+1)²)’=6.(x²+1)²+6x.(2(x²+1).(x²+1)’)=6.(x²+1)²+6x.(2(x²+1).2x)=6.(x²+1)²+24x²(x²+1)y”’=(6(x²+1)²)’+(24x²(x²+1))’=6((x²+1)²)’+(24x²)'(x²+1)+24x²(x²+1)’=6(2(x²+1)(x²+1)’)+48x(x²+1)+24x².2x=6(2(x²+1)2x)+48x(x²+1)+48x³=24x(x²+1)+48x(x²+1)+48x³=72x(x²+1)+48x³
Đáp án:
\(y”’ = 120{x^3} + 72x\)
Giải thích các bước giải:
\(\begin{array}{l}
y = {\left( {{x^2} + 1} \right)^3}\\
y’ = 3.2x.{\left( {{x^2} + 1} \right)^2} = 6x{\left( {{x^2} + 1} \right)^2}\\
y” = 6{\left( {{x^2} + 1} \right)^2} + 2.2x\left( {{x^2} + 1} \right).6x\\
= 6{\left( {{x^2} + 1} \right)^2} + 24{x^2}\left( {{x^2} + 1} \right)\\
= \left( {{x^2} + 1} \right)\left( {6{x^2} + 6 + 24{x^2}} \right)\\
= \left( {{x^2} + 1} \right)\left( {30{x^2} + 6} \right)\\
y”’ = 2x\left( {30{x^2} + 6} \right) + 30.2.x\left( {{x^2} + 1} \right)\\
= 60{x^3} + 12x + 60{x^3} + 60x\\
= 120{x^3} + 72x
\end{array}\)