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) \(y = \frac{x^{2}+7x+3}{x^{2}-3x}\).
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) \( y’=\frac{(x^{2}+7x+3)’.(x^{2}-3x)-(x^{2}+7x+3).(x^{2}-3x)’}{(x^{2}-3x)^{2}}\) = \( \frac{(2x-7).(x^{2}-3x)-(x^{2}+7x+3).(2x-3)}{(x^{2}-3x)^{2}}\) = \( \frac{-10x^{2}-6x+9}{(x^{2}-3x)^{2}}\).
Bài 2: Giải các bất phương trình sau:
a) \(y'<0\) với \({{{x^2} + x + 2} \over {x – 1}}\)
b) \(y’≥0\) với \(y = \frac{x^{2}+3}{x+1}\);
c) \(y’>0\) với \(y = \frac{2x-1}{x^{2}+x+4}\).
a) Ta có \( y’=\frac{(x^{2}+x+2)’.(x-1)-(x^{2}+x+2).(x-1)’}{(x-1)^{2}}\) = \( \frac{x^{2}-2x-3}{(x-1)^{2}}\)
Do đó, \(y'<0\Leftrightarrow \frac{x^{2}-2x-3}{(x-1)^{2}}\)
\( \Leftrightarrow \left\{ \matrix{
x \ne 1 \hfill \cr
– 1 < x < 3 \hfill \cr} \right. \Leftrightarrow \)\(x∈ (-1;1) ∪ (1;3)\).
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b) Ta có \( y’=\frac{(x^{2}+3)’.(x+1)-(x^{2}+3).(x+1)’}{(x+1)^{2}}\) = \( \frac{x^{2}+2x-3}{(x+1)^{2}}\).
Do đó, \(y’≥0 \Leftrightarrow \frac{x^{2}+2x-3}{(x+1)^{2}}≥0 \)
\( \Leftrightarrow \left\{ \matrix{
x \ne – 1 \hfill \cr
\left[ \matrix{
x \ge 1 \hfill \cr
x \le – 3 \hfill \cr} \right. \hfill \cr} \right. \Leftrightarrow \left[ \matrix{
x \ge 1 \hfill \cr
x \le – 3 \hfill \cr} \right. \Leftrightarrow x∈ (-∞;-3] ∪ [1;+∞)\).
c).Ta có \( y’=\frac{(2x-1)’.(x^{2}+x+4)-(2x-1).(x^{2}+x+4)’}{(x^{2}+x+4)}=\frac{-2x^{2}+2x+9}{(x^{2}+x+4)}\).
Do đó, \(y’>0 \Leftrightarrow \frac{-2x^{2}+2x+9}{(x^{2}+x+4)} >0\Leftrightarrow -2x^2+2x +9>0 \)\(\Leftrightarrow \frac{1-\sqrt{19}}{2} < x < \frac{1+\sqrt{19}}{2}\Leftrightarrow x∈ \left ( \frac{1-\sqrt{19}}{2};\frac{1+\sqrt{19}}{2} \right )\)
Vì \(x^2+x +4 =\) \( \left ( x+\frac{1}{2} \right )^{2}\)+ \( \frac{15}{4} >0\), với \(∀ x ∈ \mathbb R\).
Bài 3: Tìm đạo hàm của các hàm số sau:
a) \(y = 5sinx -3cosx\);
b) \( y=\frac{sinx+cosx}{sinx-cosx}\);
c) \(y = x cotx\);
d) \(y = \frac{sinx}{x}\) + \( \frac{x}{sinx}\);
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e) \(y = \sqrt{(1 +2tan x)}\);
f) \(y = sin\sqrt{(1 +x^2)}\).
a) \(y’=5cosx-3(-sinx)=5cosx+3sinx\);
b) \( y’={{(sinx+cos x)’.(sin x- cos x)-(sin x+cos x)(sin x-cos x)’}\over{(sin x-cos x)^{2}}}\) = \( {{(cos x-sin x)(sin x -cos x)-(sin x+ cos x)(cosx+sinx)}\over{(sin x-cosx )^{2}}}\) = \( {{-2}\over{(sin x-cos x)^{2}}}\).
c) \(y’ = cotx +x. \left ( -\frac{1}{sin^{2}x} \right )= cotx – \frac{x}{sin^{2}x}\).
d) \( y’=\frac{(sin x)’.x-sin x.(x)’}{x^{2}}\) +\( \frac{(x)’.sin x-x(sin x)’}{sin^{2}x}\) = \( \frac{x.cosx-sinx}{x^{2}}+\frac{sin x-x.cosx}{sin^{2}x}\)\( = (x. cosx -sinx) \left ( \frac{1}{x^{2}}-\frac{1}{sin^{2}x} \right )\).
e) \( y’=\frac{(1+2tanx)’}{2\sqrt{1+2tanx}}\) = \( \frac{\frac{2}{cos^{2}x}}{2\sqrt{1+2tanx}}\) = \( \frac{1}{cos^{2}x\sqrt{1+2tanx}}\).
f) \(y’ = (\sqrt{(1+x^2)})’ cos\sqrt{(1+x^2)} \)\(= \frac{(1+x^{2})’}{2\sqrt{1+x^{2}}}cos\sqrt{(1+x^2)} = \frac{x}{\sqrt{1+x^{2}}}cos\sqrt{(1+x^2)}\).
Bài 4: 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) \(y = cos\frac{x}{1+x}\).
a) \(y’ = \left( {9 – 2x} \right)'(2{x^3} – 9{x^2} + 1) + \left( {9 – 2x} \right)(2{x^3} – 9{x^2} + 1)’\)
\(= – 2(2{x^3} – 9{x^2} + 1) + \left( {9 – 2x} \right)(6{x^2} – 18x) \)
\(= – 16{x^3} + 108{x^2} – 162x – 2\).
b) \(y’ = \left ( 6\sqrt{x} -\frac{1}{x^{2}}\right )’.(7x -3) +\left ( 6\sqrt{x} -\frac{1}{x^{2}}\right )(7x -3)’\)
\(= \left ( \frac{3}{\sqrt{x}} +\frac{2}{x^{3}}\right )(7x -3) +7 \left ( 6\sqrt{x} -\frac{1}{x^{2}}\right )\).
c) \(y’ = (x -2)’\sqrt{(x^2+1)} + (x -2)\sqrt {(x^2+1)}’ \)
\(= \sqrt {(x^2+1)} + (x -2)\frac{\left ( x^{2}+1 \right )’}{2\sqrt{x^{2}+1}}\)
\(= \sqrt {(x^2+1)} + (x -2) \frac{2x}{2\sqrt{x^{2}+1}}\)
\( = \sqrt {(x^2+1)} + \frac{x^{2}-2x}{\sqrt{x^{2}+1}}\) = \( \frac{2x^{2}-2x+1}{\sqrt{x^{2}+1}}\).
d) \(y’ = 2tanx.(tanx)’ – (x^2)’ \left ( -\frac{1}{sin^{2}x^{2}} \right )\) = \( \frac{2tanx}{cos^{2}x}+\frac{2x}{sin^{2}x^{2}}\).
e) \(y’ = \left ( \frac{1}{1+x} \right )’sin \frac{x}{1+x}\) = \( -\frac{1}{(1+x)^{2}}sin \frac{x}{1+x}\).
