Giải câu 3 bài 2: Quy tắc tính đạo hàm

a) \(y = {({x^{7}} - 5{x^2})^3}\)

\(y' = 3.{({x^7} - 5{x^2})^2}.({x^7} - 5{x^2})' \)

\(= 3.{({x^{7}} - 5{x^2})^2}.(7{x^6} - 10x)\)

\(= 3x^5.{({x^{5}} - 5)^2}(7{x^5} - 10)\)

b) \(y = ({x^2} + 1)(5 - 3{x^2})\)

\(= 5{x^2} - 3{x^4} + 5 - 3{x^2} \)

\(=  - 3{x^4} + 2{x^2} + 5\)

\(y' =  - 12{x^3} + 4x =  - 4x.(3{x^2} - 1)\).

c) \(y =  \frac{2x}{x^{2}-1}\)

\(y' = \frac{\left ( 2x \right )'.\left ( x^{2}-1 \right )-2x\left ( x^{2}-1 \right )'}{\left ( x^{2}-1 \right )^{2}}\)

\(= \frac{2.\left ( x^{2}-1 \right )-2x.2x}{\left ( x^{2}-1 \right )^{2}}\)

\(= \frac{-2\left ( x^{2}+1 \right )}{\left ( x^{2}-1 \right )^{2}}\)

d) \(y =  \frac{3-5x}{x^{2}-x+1}\)

\(y' =  \frac{\left ( 3-5x \right )'\left ( x^{2}-x+1 \right )-\left ( 3-5x \right ).\left ( x^{2}-x+1 \right )'}{\left ( x^{2}-x+1 \right )^{2}}\)

\(= \frac{-5\left ( x^{2}-x+1 \right )-\left ( 3-5x \right ).\left ( 2x-1 \right )}{\left ( x^{2}-x+1 \right )^{2}}\)

\(= \frac{-5x^2+5x-5-6x+3+10x^2-5x}{( x^{2}-x+1)^{2}}\)

\(= \frac{5x^{2}-6x-2}{\left ( x^{2}-x+1 \right )^{2}}\).

e) \(y = \left ( m+\frac{n}{x^{2}} \right )^{3}\) (\(m, n\) là các hằng số)

\(y' = 3. \left ( m+\frac{n}{x^{2}} \right )^{2} . \left ( m+\frac{n}{x^{2}} \right )'\)

\(=3. \left ( m+\frac{n}{x^{2}} \right )^{2} \left ( -\frac{2n}{x^{3}} \right )\)

\(=- \frac{6n}{x^{3}} . \left ( m+\frac{n}{x^{2}} \right )^{2}\).