Giải bài 1 trang 168 – SGK môn Đại số và Giải tích lớp 11
Tìm đạo hàm của các hàm số sau
\(a)\, y=\dfrac{x-1}{5x-2} \)
\(b)\, y=\dfrac{2x+3}{7-3x} \)
\(c)\, y=\dfrac{{{x}^{2}}+2x+3}{3-4x} \)
\(d)\, y=\dfrac{{{x}^{2}}+7x+3}{{{x}^{2}}-3x} \)
Gợi ý:
Sử dụng công thức \(\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^2}\)
a)
\(\begin{align} & y'=\dfrac{\left( x-1 \right)'\left( 5x-2 \right)-\left( x-1 \right)\left( 5x-2 \right)'}{{{\left( 5x-2 \right)}^{2}}} \\ & =\dfrac{\left( 5x-2 \right)-5\left( x-1 \right)}{{{\left( 5x-2 \right)}^{2}}}=\dfrac{3}{{{\left( 5x-2 \right)}^{2}}} \\ \end{align} \)
b)
\(\begin{align} & y'=\dfrac{\left( 2x+3 \right)'\left( 7-3x \right)-\left( 2x+3 \right)\left( 7-3x \right)'}{{{\left( 7-3x \right)}^{2}}} \\ & =\dfrac{2\left( 7-3x \right)+3\left( 2x+3 \right)}{{{\left( 7-3x \right)}^{2}}} \\ & =\dfrac{23}{{{\left( 7-3x \right)}^{2}}} \\ \end{align} \)
c)
\(\begin{align} & y'=\dfrac{\left( {{x}^{2}}+2x+3 \right)'\left( 3-4x \right)-\left( {{x}^{2}}+2x+3 \right)\left( 3-4x \right)'}{{{\left( 3-4x \right)}^{2}}} \\ & =\dfrac{\left( 2x+2 \right)\left( 3-4x \right)-\left( {{x}^{2}}+2x+3 \right)\left( -4 \right)}{{{\left( 3-4x \right)}^{2}}} \\ & =\dfrac{-4{{x}^{2}}+6x+18}{{{\left( 3-4x \right)}^{2}}} \\ \end{align} \)
d)
\(\begin{align} & y'=\dfrac{\left( {{x}^{2}}+7x+3 \right)'\left( {{x}^{2}}-3x \right)-\left( {{x}^{2}}+7x+3 \right)\left( {{x}^{2}}-3x \right)'}{{{\left( {{x}^{2}}-3x \right)}^{2}}} \\ & =\dfrac{\left( 2x+7 \right)\left( {{x}^{2}}-3x \right)-\left( {{x}^{2}}+7x+3 \right)\left( 2x-3 \right)}{{{\left( {{x}^{2}}-3x \right)}^{2}}} \\ & =\dfrac{-10{{x}^{2}}-6x+9}{{{\left( {{x}^{2}}-3x \right)}^{2}}} \\ \end{align} \)