Câu 4: trang 169 sgk toán Đại số và giải tích 11
Tìm đạo hàm của các hàm số sau:
a) \(y = \left( {9 - 2x} \right)(2{x^3} - 9{x^2} + 1)\)
b) \(y = \left ( 6\sqrt{x} -\frac{1}{x^{2}}\right )(7x -3)\)
c) \(y = (x -2)\sqrt{(x^2+1)}\)
d) \(y = tan^2x +cotx^2\)
e) \(y = cos\frac{x}{1+x}\)
Bài Làm:
a) \(y = \left( {9 - 2x} \right)(2{x^3} - 9{x^2} + 1)\)
\(y' = \left( {9 - 2x} \right)'(2{x^3} - 9{x^2} + 1) + \left( {9 - 2x} \right)(2{x^3} - 9{x^2} + 1)'\)
\(= - 2(2{x^3} - 9{x^2} + 1) + \left( {9 - 2x} \right)(6{x^2} - 18x) \)
\(=-4x^3+18x^2-2+54x^2-162x-12x^3+36x^2\)
\(= - 16{x^3} + 108{x^2} - 162x - 2\).
b) \(y = \left ( 6\sqrt{x} -\frac{1}{x^{2}}\right )(7x -3)\)
\(y' = \left ( 6\sqrt{x} -\frac{1}{x^{2}}\right )'.(7x -3) +\left ( 6\sqrt{x} -\frac{1}{x^{2}}\right )(7x -3)'\)
\(= \left ( \frac{3}{\sqrt{x}} +\frac{2}{x^{3}}\right )(7x -3) +7 \left ( 6\sqrt{x} -\frac{1}{x^{2}}\right )\)
\(=63\sqrt{x}-\frac{9}{\sqrt{x}}+\frac{7}{x^2}-\frac{6}{x^3}\)
c) \(y = (x -2)\sqrt{(x^2+1)}\)
\(y' = (x -2)'\sqrt{(x^2+1)} + (x -2)\sqrt {(x^2+1)}' \)
\(= \sqrt {(x^2+1)} + (x -2)\frac{\left ( x^{2}+1 \right )'}{2\sqrt{x^{2}+1}}\)
\(= \sqrt {(x^2+1)} + (x -2) \frac{2x}{2\sqrt{x^{2}+1}}\)
\( = \sqrt {(x^2+1)} + \frac{x^{2}-2x}{\sqrt{x^{2}+1}}\)
\(= \frac{2x^{2}-2x+1}{\sqrt{x^{2}+1}}\).
d) \(y = tan^2x +cotx^2\)
\(y' = 2tanx.(tanx)' - (x^2)' \left ( -\frac{1}{sin^{2}x^{2}} \right )= \frac{2tanx}{cos^{2}x}+\frac{2x}{sin^{2}x^{2}}\)
e) \(y = cos\frac{x}{1+x}\)
\(y' = \left ( \frac{1}{1+x} \right )'sin \frac{x}{1+x}= -\frac{1}{(1+x)^{2}}sin \frac{x}{1+x}\).