Bài 2: Quy tắc tính đạo hàm

\(y=\left(x^7+x\right)^3\)

\(y'=3\left(x^7+x\right)^2.\left(x^7+x\right)'=3\left(x^7+x\right)^2\left(7x^6+1\right)\)

y'=6(x+1)^5.(x+1)'=6(x+1)^5

d.

\(y'=12x^2-1\)

e.

\(y'=\dfrac{\left(x-1\right)'\left(3x+1\right)-\left(3x+1\right)'\left(x-1\right)}{\left(3x+1\right)^2}=\dfrac{3x+1-3\left(x-1\right)}{\left(3x+1\right)^2}=\dfrac{4}{\left(3x+1\right)^2}\)

i.

\(y'=15x^2+\dfrac{1}{2\sqrt{x}}+\dfrac{12}{x^2}\)

a.

\(y=x^7\left(x+1\right)=x^8+x^7\)

\(\Rightarrow y'=8x^7+7x^6\)

b.

\(y'=2x+\dfrac{2}{2\sqrt{x}}=2x+\dfrac{1}{\sqrt{x}}\)

c.

\(y'=9x^2+4x\)

\(y'=-3.\left(\sqrt{x^2+2}\right)'.\dfrac{-1}{\left(\sqrt{x^2+2}\right)^2}=3.\dfrac{\left(x^2+2\right)'}{2\sqrt{x^2+2}}.\dfrac{1}{x^2+2}=\dfrac{3x}{\left(x^2+2\right)\sqrt{x^2+2}}\)

\(y'=\left(\dfrac{2x-1}{x+1}\right)'.\dfrac{1}{2\sqrt{\dfrac{2x-1}{x+1}}}=\dfrac{3}{2\left(x+1\right)^2}.\sqrt{\dfrac{x+1}{2x-1}}\)

\(y'=\left(3x^2+4\right)'\sqrt{x}+\left(3x^2+4\right).\left(\sqrt{x}\right)'=6x\sqrt{x}+\dfrac{3x^2+4}{2\sqrt{x}}=\dfrac{15x^2+4}{2\sqrt{x}}\)

Đề bài yêu cầu gì bạn nhỉ?

\(f\left(x+3\right)=g\left(x\right)+x^2-10x+5\)

\(\Rightarrow f'\left(x+3\right)=g'\left(x\right)+2x-10\)

Thế \(x=1\) ta được:

\(f'\left(4\right)=g'\left(1\right)-8\)

\(\Rightarrow g'\left(1\right)=f'\left(4\right)+8=13\)

Không nhìn rõ đề, chắc là 2 câu tính đạo hàm?

a. \(y'=2x^3+6x\)

b.

\(y'=\dfrac{\left(4x-6\right)\left(x+4\right)-\left(2x^2-6x+5\right)}{\left(x+4\right)^2}=\dfrac{2x^2+16x-29}{\left(x+4\right)^2}\)