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