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Câu hỏi: Tính đạo hàm của các hàm số sau:
Lời giải chi tiết:
$y=\dfrac{1+x-x^{2}}{1-x+x^{2}}$
$y^{\prime}=\dfrac{\left(1+x-x^{2}\right)^{\prime}\left(1-x+x^{2}\right)-\left(1+x-x^{2}\right)\left(1-x+x^{2}\right)}{\left(1-x+x^{2}\right)^{2}}$
$=\dfrac{(1-2 x)\left(1-x+x^{2}\right)-\left(1+x-x^{2}\right)(-1+2 x)}{\left(1-x+x^{2}\right)^{2}}$
$=\dfrac{(1-2 x)\left(1-x+x^{2}\right)+\left(1+x-x^{2}\right)(1-2 x)}{\left(1-x+x^{2}\right)^{2}}$
$=\dfrac{2(1-2 x)}{\left(1-x+x^{2}\right)^{2}}$
Lời giải chi tiết:
$y=\dfrac{\left(2-x^{2}\right)\left(3-x^{3}\right)}{(1-x)^{2}}=\dfrac{6-3 x^{2}-2 x^{3}+x^{5}}{1-2 x+x^{2}}$
$y^{\prime}=\dfrac{\left(6-3 x^{2}-2 x^{3}+x^{5}\right)^{\prime}\left(1-2 x+x^{2}\right)-\left(6-3 x^{2}-2 x^{3}+x^{5}\right)\left(1-2 x+x^{2}\right)^{3}}{\left(1-2 x+x^{2}\right)^{2}}$
$=\dfrac{\left(-6 x-6 x^{2}+5 x^{4}\right)\left(1-2 x+x^{2}\right)-\left(6-3 x^{2}-2 x^{3}+x^{5}\right)(-2+2 x)}{\left(1-2 x+x^{2}\right)^{2}}$
$=\dfrac{\left(-6 x-6 x^{2}+5 x^{4}\right)(x-1)^{2}-2\left(6-3 x^{2}-2 x^{3}+x^{5}\right)(x-1)}{\left[(x-1)^{2}\right]^{2}}$
$=\dfrac{(x-1)\left[\left(-6 x-6 x^{2}+5 x^{4}\right)(x-1)-2\left(6-3 x^{2}-2 x^{3}+x^{5}\right)\right]}{(x-1)^{4}}$
$=\dfrac{-12+6 x+6 x^{2}-2 x^{3}-5 x^{4}+3 x^{5}}{(x-1)^{3}}$
Lời giải chi tiết:
\(\begin{array}{l}y = \cos 2x - 2\sin x\\y' = \left( {\cos 2x} \right)' - 2\left({\sin x} \right)'\\ = - \left({2x} \right)'\sin 2x - 2\cos x\\ = - 2\sin 2x - 2\cos x\end{array}\)
Lời giải chi tiết:
\(\begin{array}{l}y = \dfrac{{\cos x}}{{2{{\sin }^2}x}}\\y' = \dfrac{{\left( {\cos x} \right)'. 2{{\sin }^2}x - \cos x\left({2{{\sin }^2}x} \right)'}}{{4{{\sin }^4}x}}\\ = \dfrac{{ - \sin x. 2{{\sin }^2}x - \cos x. 2.2\left({\sin x} \right)'\sin x}}{{4{{\sin }^4}x}}\\ = \dfrac{{ - 2{{\sin }^3}x - 4\cos x.\cos x.\sin x}}{{4{{\sin }^4}x}}\\ = \dfrac{{ - 2\sin x\left({{{\sin }^2}x + 2{{\cos }^2}x} \right)}}{{4{{\sin }^4}x}}\\ = - \dfrac{{{{\sin }^2}x + {{\cos }^2}x + {{\cos }^2}x}}{{2{{\sin }^3}x}}\\ = - \dfrac{{1 + {{\cos }^2}x}}{{2{{\sin }^3}x}}\end{array}\)
Lời giải chi tiết:
\(y = {\cos ^2}\dfrac{x}{3}\tan \dfrac{x}{2}\)
\(\begin{array}{l}y' = \left( {{{\cos }^2}\dfrac{x}{3}} \right)'\tan \dfrac{x}{2} + {\cos ^2}\dfrac{x}{3}\left({\tan \dfrac{x}{2}} \right)'\\ = 2\cos \dfrac{x}{3}.\left({\cos \dfrac{x}{3}} \right)'.\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} + {\cos ^2}\dfrac{x}{3}.\dfrac{{\left({\dfrac{x}{2}} \right)'}}{{{{\cos }^2}\dfrac{x}{2}}}\\ = 2\cos \dfrac{x}{3}.\left({\dfrac{x}{3}} \right)'.\left({ - \sin \dfrac{x}{3}} \right).\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} + {\cos ^2}\dfrac{x}{3}.\dfrac{{\dfrac{1}{2}}}{{{{\cos }^2}\dfrac{x}{2}}}\\ = - 2\cos \dfrac{x}{3}.\dfrac{1}{3}\sin \dfrac{x}{3}.\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} + \dfrac{1}{2}.\dfrac{{{{\cos }^2}\dfrac{x}{3}}}{{{{\cos }^2}\dfrac{x}{2}}}\\ = - \dfrac{1}{3}\sin \dfrac{{2x}}{3}\tan \dfrac{x}{2} + \dfrac{{{{\cos }^2}\dfrac{x}{3}}}{{2{{\cos }^2}\dfrac{x}{2}}}\end{array}\)
Lời giải chi tiết:
\(\begin{array}{l}y = \sqrt {\sin \left( {2x - \dfrac{\pi }{6}} \right)} \\y' = \dfrac{{\left[ {\sin \left({2x - \dfrac{\pi }{6}} \right)} \right]'}}{{2\sqrt {\sin \left({2x - \dfrac{\pi }{6}} \right)} }}\\ = \dfrac{{\left({2x - \dfrac{\pi }{6}} \right)'.\cos \left({2x - \dfrac{\pi }{6}} \right)}}{{2\sqrt {\sin \left({2x - \dfrac{\pi }{6}} \right)} }}\\ = \dfrac{{2\cos \left({2x - \dfrac{\pi }{6}} \right)}}{{2\sqrt {\sin \left({2x - \dfrac{\pi }{6}} \right)} }}\\ = \dfrac{{\cos \left({2x - \dfrac{\pi }{6}} \right)}}{{\sqrt {\sin \left({2x - \dfrac{\pi }{6}} \right)} }}\end{array}\)
Lời giải chi tiết:
\(y = \cos \dfrac{x}{{x + 1}}\)
\(\begin{array}{l}y' = \left( {\dfrac{x}{{x + 1}}} \right)'.\left({ - \sin \dfrac{x}{{x + 1}}} \right)\\ = \dfrac{{\left(x \right)'\left({x + 1} \right) - x\left({x + 1} \right)'}}{{{{\left({x + 1} \right)}^2}}}.\left({ - \sin \dfrac{x}{{x + 1}}} \right)\\ = - \dfrac{{1.\left({x + 1} \right) - x. 1}}{{{{\left({x + 1} \right)}^2}}}\sin \dfrac{x}{{x + 1}}\\ = - \dfrac{1}{{{{\left({x + 1} \right)}^2}}}\sin \dfrac{x}{{x + 1}}\end{array}\)
Lời giải chi tiết:
\(y = \dfrac{{{x^2} - 1}}{{\sin 3x}}\)
\(\begin{array}{l}y' = \dfrac{{\left( {{x^2} - 1} \right)'\sin 3x - \left({{x^2} - 1} \right).\left({\sin 3x} \right)'}}{{{{\sin }^2}3x}}\\ = \dfrac{{2x\sin 3x - \left({{x^2} - 1} \right).\left({3x} \right)'\cos 3x}}{{{{\sin }^2}3x}}\\ = \dfrac{{2x\sin 3x - \left({{x^2} - 1} \right). 3\cos 3x}}{{{{\sin }^2}3x}}\\ = \dfrac{{2x\sin 3x - 3\left({{x^2} - 1} \right)\cos 3x}}{{{{\sin }^2}3x}}\end{array}\)
Lời giải chi tiết:
\(\begin{array}{l}y = 3{\sin ^2}x\cos x + {\cos ^2}x\\y' = 3.\left[ {\left( {{{\sin }^2}x} \right)'\cos x + {{\sin }^2}x\left({\cos x} \right)'} \right] + 2\cos x\left({\cos x} \right)'\\ = 3\left[ {2\sin x\left({\sin x} \right)'\cos x + {{\sin }^2}x.\left({ - \sin x} \right)} \right] + 2\cos x\left({ - \sin x} \right)\\ = 3\left({2\sin x\cos x\cos x - {{\sin }^3}x} \right) - 2\sin x\cos x\\ = 3\left({\sin 2x\cos x - {{\sin }^3}x} \right) - \sin 2x\\ = 3\sin 2x\cos x - 3{\sin ^3}x - \sin 2x\\ = \sin 2x\left({3\cos x - 1} \right) - 3{\sin ^3}x\end{array}\)
Lời giải chi tiết:
\(\begin{array}{l}y = \sqrt {7 - 4x} \cot 3x\\y' = \left( {\sqrt {7 - 4x} } \right)'\cot 3x + \sqrt {7 - 4x} \left({\cot 3x} \right)'\\ = \dfrac{{\left({7 - 4x} \right)'}}{{2\sqrt {7 - 4x} }}.\cot 3x + \sqrt {7 - 4x} .\dfrac{{ - \left({3x} \right)'}}{{{{\sin }^2}3x}}\\ = \dfrac{{ - 4}}{{2\sqrt {7 - 4x} }}.\cot 3x + \sqrt {7 - 4x} .\dfrac{{ - 3}}{{{{\sin }^2}3x}}\\ = \dfrac{{ - 2\cot 3x}}{{\sqrt {7 - 4x} }} - \dfrac{{3\sqrt {7 - 4x} }}{{{{\sin }^2}3x}}\end{array}\)
Câu a
\(y = \dfrac{{1 + x - {x^2}}}{{1 - x + {x^2}}}\)Lời giải chi tiết:
$y=\dfrac{1+x-x^{2}}{1-x+x^{2}}$
$y^{\prime}=\dfrac{\left(1+x-x^{2}\right)^{\prime}\left(1-x+x^{2}\right)-\left(1+x-x^{2}\right)\left(1-x+x^{2}\right)}{\left(1-x+x^{2}\right)^{2}}$
$=\dfrac{(1-2 x)\left(1-x+x^{2}\right)-\left(1+x-x^{2}\right)(-1+2 x)}{\left(1-x+x^{2}\right)^{2}}$
$=\dfrac{(1-2 x)\left(1-x+x^{2}\right)+\left(1+x-x^{2}\right)(1-2 x)}{\left(1-x+x^{2}\right)^{2}}$
$=\dfrac{2(1-2 x)}{\left(1-x+x^{2}\right)^{2}}$
Câu b
\(y = \dfrac{{\left( {2 - {x^2}} \right)\left({3 - {x^3}} \right)}}{{{{\left({1 - x} \right)}^2}}}\)Lời giải chi tiết:
$y=\dfrac{\left(2-x^{2}\right)\left(3-x^{3}\right)}{(1-x)^{2}}=\dfrac{6-3 x^{2}-2 x^{3}+x^{5}}{1-2 x+x^{2}}$
$y^{\prime}=\dfrac{\left(6-3 x^{2}-2 x^{3}+x^{5}\right)^{\prime}\left(1-2 x+x^{2}\right)-\left(6-3 x^{2}-2 x^{3}+x^{5}\right)\left(1-2 x+x^{2}\right)^{3}}{\left(1-2 x+x^{2}\right)^{2}}$
$=\dfrac{\left(-6 x-6 x^{2}+5 x^{4}\right)\left(1-2 x+x^{2}\right)-\left(6-3 x^{2}-2 x^{3}+x^{5}\right)(-2+2 x)}{\left(1-2 x+x^{2}\right)^{2}}$
$=\dfrac{\left(-6 x-6 x^{2}+5 x^{4}\right)(x-1)^{2}-2\left(6-3 x^{2}-2 x^{3}+x^{5}\right)(x-1)}{\left[(x-1)^{2}\right]^{2}}$
$=\dfrac{(x-1)\left[\left(-6 x-6 x^{2}+5 x^{4}\right)(x-1)-2\left(6-3 x^{2}-2 x^{3}+x^{5}\right)\right]}{(x-1)^{4}}$
$=\dfrac{-12+6 x+6 x^{2}-2 x^{3}-5 x^{4}+3 x^{5}}{(x-1)^{3}}$
Câu c
\(y = \cos 2x - 2\sin x\)Lời giải chi tiết:
\(\begin{array}{l}y = \cos 2x - 2\sin x\\y' = \left( {\cos 2x} \right)' - 2\left({\sin x} \right)'\\ = - \left({2x} \right)'\sin 2x - 2\cos x\\ = - 2\sin 2x - 2\cos x\end{array}\)
Câu d
\(y = \dfrac{{\cos x}}{{2{{\sin }^2}x}}\)Lời giải chi tiết:
\(\begin{array}{l}y = \dfrac{{\cos x}}{{2{{\sin }^2}x}}\\y' = \dfrac{{\left( {\cos x} \right)'. 2{{\sin }^2}x - \cos x\left({2{{\sin }^2}x} \right)'}}{{4{{\sin }^4}x}}\\ = \dfrac{{ - \sin x. 2{{\sin }^2}x - \cos x. 2.2\left({\sin x} \right)'\sin x}}{{4{{\sin }^4}x}}\\ = \dfrac{{ - 2{{\sin }^3}x - 4\cos x.\cos x.\sin x}}{{4{{\sin }^4}x}}\\ = \dfrac{{ - 2\sin x\left({{{\sin }^2}x + 2{{\cos }^2}x} \right)}}{{4{{\sin }^4}x}}\\ = - \dfrac{{{{\sin }^2}x + {{\cos }^2}x + {{\cos }^2}x}}{{2{{\sin }^3}x}}\\ = - \dfrac{{1 + {{\cos }^2}x}}{{2{{\sin }^3}x}}\end{array}\)
Câu e
\(y = {\cos ^2}\dfrac{x}{3}\tan \dfrac{x}{2}\)Lời giải chi tiết:
\(y = {\cos ^2}\dfrac{x}{3}\tan \dfrac{x}{2}\)
\(\begin{array}{l}y' = \left( {{{\cos }^2}\dfrac{x}{3}} \right)'\tan \dfrac{x}{2} + {\cos ^2}\dfrac{x}{3}\left({\tan \dfrac{x}{2}} \right)'\\ = 2\cos \dfrac{x}{3}.\left({\cos \dfrac{x}{3}} \right)'.\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} + {\cos ^2}\dfrac{x}{3}.\dfrac{{\left({\dfrac{x}{2}} \right)'}}{{{{\cos }^2}\dfrac{x}{2}}}\\ = 2\cos \dfrac{x}{3}.\left({\dfrac{x}{3}} \right)'.\left({ - \sin \dfrac{x}{3}} \right).\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} + {\cos ^2}\dfrac{x}{3}.\dfrac{{\dfrac{1}{2}}}{{{{\cos }^2}\dfrac{x}{2}}}\\ = - 2\cos \dfrac{x}{3}.\dfrac{1}{3}\sin \dfrac{x}{3}.\dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}} + \dfrac{1}{2}.\dfrac{{{{\cos }^2}\dfrac{x}{3}}}{{{{\cos }^2}\dfrac{x}{2}}}\\ = - \dfrac{1}{3}\sin \dfrac{{2x}}{3}\tan \dfrac{x}{2} + \dfrac{{{{\cos }^2}\dfrac{x}{3}}}{{2{{\cos }^2}\dfrac{x}{2}}}\end{array}\)
Câu f
\(y = \sqrt {\sin \left( {2x - \dfrac{\pi }{6}} \right)}\)Lời giải chi tiết:
\(\begin{array}{l}y = \sqrt {\sin \left( {2x - \dfrac{\pi }{6}} \right)} \\y' = \dfrac{{\left[ {\sin \left({2x - \dfrac{\pi }{6}} \right)} \right]'}}{{2\sqrt {\sin \left({2x - \dfrac{\pi }{6}} \right)} }}\\ = \dfrac{{\left({2x - \dfrac{\pi }{6}} \right)'.\cos \left({2x - \dfrac{\pi }{6}} \right)}}{{2\sqrt {\sin \left({2x - \dfrac{\pi }{6}} \right)} }}\\ = \dfrac{{2\cos \left({2x - \dfrac{\pi }{6}} \right)}}{{2\sqrt {\sin \left({2x - \dfrac{\pi }{6}} \right)} }}\\ = \dfrac{{\cos \left({2x - \dfrac{\pi }{6}} \right)}}{{\sqrt {\sin \left({2x - \dfrac{\pi }{6}} \right)} }}\end{array}\)
Câu g
\(y = \cos \dfrac{x}{{x + 1}}\)Lời giải chi tiết:
\(y = \cos \dfrac{x}{{x + 1}}\)
\(\begin{array}{l}y' = \left( {\dfrac{x}{{x + 1}}} \right)'.\left({ - \sin \dfrac{x}{{x + 1}}} \right)\\ = \dfrac{{\left(x \right)'\left({x + 1} \right) - x\left({x + 1} \right)'}}{{{{\left({x + 1} \right)}^2}}}.\left({ - \sin \dfrac{x}{{x + 1}}} \right)\\ = - \dfrac{{1.\left({x + 1} \right) - x. 1}}{{{{\left({x + 1} \right)}^2}}}\sin \dfrac{x}{{x + 1}}\\ = - \dfrac{1}{{{{\left({x + 1} \right)}^2}}}\sin \dfrac{x}{{x + 1}}\end{array}\)
Câu h
\(y = \dfrac{{{x^2} - 1}}{{\sin 3x}}\)Lời giải chi tiết:
\(y = \dfrac{{{x^2} - 1}}{{\sin 3x}}\)
\(\begin{array}{l}y' = \dfrac{{\left( {{x^2} - 1} \right)'\sin 3x - \left({{x^2} - 1} \right).\left({\sin 3x} \right)'}}{{{{\sin }^2}3x}}\\ = \dfrac{{2x\sin 3x - \left({{x^2} - 1} \right).\left({3x} \right)'\cos 3x}}{{{{\sin }^2}3x}}\\ = \dfrac{{2x\sin 3x - \left({{x^2} - 1} \right). 3\cos 3x}}{{{{\sin }^2}3x}}\\ = \dfrac{{2x\sin 3x - 3\left({{x^2} - 1} \right)\cos 3x}}{{{{\sin }^2}3x}}\end{array}\)
Câu i
\(y = 3{\sin ^2}x\cos x + {\cos ^2}x\)Lời giải chi tiết:
\(\begin{array}{l}y = 3{\sin ^2}x\cos x + {\cos ^2}x\\y' = 3.\left[ {\left( {{{\sin }^2}x} \right)'\cos x + {{\sin }^2}x\left({\cos x} \right)'} \right] + 2\cos x\left({\cos x} \right)'\\ = 3\left[ {2\sin x\left({\sin x} \right)'\cos x + {{\sin }^2}x.\left({ - \sin x} \right)} \right] + 2\cos x\left({ - \sin x} \right)\\ = 3\left({2\sin x\cos x\cos x - {{\sin }^3}x} \right) - 2\sin x\cos x\\ = 3\left({\sin 2x\cos x - {{\sin }^3}x} \right) - \sin 2x\\ = 3\sin 2x\cos x - 3{\sin ^3}x - \sin 2x\\ = \sin 2x\left({3\cos x - 1} \right) - 3{\sin ^3}x\end{array}\)
Câu k
\(y = \sqrt {7 - 4x} \cot 3x\)Lời giải chi tiết:
\(\begin{array}{l}y = \sqrt {7 - 4x} \cot 3x\\y' = \left( {\sqrt {7 - 4x} } \right)'\cot 3x + \sqrt {7 - 4x} \left({\cot 3x} \right)'\\ = \dfrac{{\left({7 - 4x} \right)'}}{{2\sqrt {7 - 4x} }}.\cot 3x + \sqrt {7 - 4x} .\dfrac{{ - \left({3x} \right)'}}{{{{\sin }^2}3x}}\\ = \dfrac{{ - 4}}{{2\sqrt {7 - 4x} }}.\cot 3x + \sqrt {7 - 4x} .\dfrac{{ - 3}}{{{{\sin }^2}3x}}\\ = \dfrac{{ - 2\cot 3x}}{{\sqrt {7 - 4x} }} - \dfrac{{3\sqrt {7 - 4x} }}{{{{\sin }^2}3x}}\end{array}\)
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