Question

[nanako]

Tính đạo hàm:

a) y= \(\left(x^5+2x\right).\left(x^6-3\right).\left(3x^7+6x^2-2\right)\)

b) y= \(\left(x^4-\dfrac{2}{3x}\right)^5\)tại x=10

c) y= \(\dfrac{5x-2}{x+1}\) tại x=4

Answer 1

[Nguyễn Việt Lâm]

a. Làm gọn 1 chút xíu:

\(y=\left(x^{11}+2x^7-3x^5-6x\right)\left(3x^7+6x^2-2\right)\)

\(y'=\left(11x^{10}+14x^6-15x^4-6\right)\left(3x^7+6x^2-2\right)+\left(21x^6+12x\right)\left(x^{11}+2x^7-3x^5-6x\right)\)

b.

 \(y'=5\left(x^4-\dfrac{2}{3x}\right)^4\left(4x^3+\dfrac{2}{3x^2}\right)\Rightarrow y'\left(10\right)=5\left(10^4-\dfrac{2}{30}\right)^4\left(4.10^3+\dfrac{2}{300}\right)=?\)

c.

\(y'=\dfrac{7}{\left(x+1\right)^2}\Rightarrow y'\left(4\right)=\dfrac{7}{25}\)