Nội dung bài giảng
Bài 1. Tìm đạo hàm của các hàm số sau:
a) \(y = \frac{x-1}{5x-2}\);
b) \(y = \frac{2x+3}{7-3x}\);
c) \(y = \frac{x^{2}+2x+3}{3-4x}\);
d) \(y = \frac{x^{2}+7x+3}{x^{2}-3x}\).
Lời giải:
a) \( y'=\frac{\left ( x-1 \right )'.\left ( 5x-2 \right )-\left ( x-1 \right ).\left ( 5x-2 \right )'}{\left ( 5x-2 \right )^{2}}\) = \( \frac{5x-2-\left ( x-1 \right ).5}{\left ( 5x-2 \right )^{2}}\) = \( \frac{3}{\left ( 5x-2 \right )^{2}}\).
b) \( y'=\frac{\left ( 2x+3 \right )'.\left ( 7-3x \right )-\left ( 2x+3 \right ).\left ( 7-3x \right )'}{\left ( 7-3x \right )^{2}}\) = \( \frac{2\left ( 7-3x \right )-\left ( 2x+3 \right ).\left ( -3 \right )}{\left ( 7-3x \right )^{2}}\) = \( \frac{23}{\left ( 7-3x \right )^{2}}\).
c) \( y'=\frac{\left ( x^{2}+2x+3 \right )'.\left ( 3-4x \right )-\left ( x^{2} +2x+3\right ).\left ( 3-4x \right )'}{\left ( 3-4x \right )^{2}}\) = \( \frac{\left ( 2x+2 \right ).\left ( 3-4x \right )-\left ( x^{2}+2x+3 \right ).(-4)}{(3-4x)^{2}}\) = \( \frac{-2(2x^{2}-3x-9)}{(3-4x)^{2}}\).
d) \( y'=\frac{(x^{2}+7x+3)'.(x^{2}-3x)-(x^{2}+7x+3).(x^{2}-3x)'}{(x^{2}-3x)^{2}}\) = \( \frac{(2x-7).(x^{2}-3x)-(x^{2}+7x+3).(2x-3)}{(x^{2}-3x)^{2}}\) = \( \frac{-10x^{2}-6x+9}{(x^{2}-3x)^{2}}\).