Giải bài 17 Ôn tập cuối năm

a. \(y' =  - {{(co{s^2}3x)'} \over {{{\cos }^4}3x}} =  - {{2\cos 3x(cos3x)'} \over {{{\cos }^4}3x}} = {{6\cos 3x\sin 3x} \over {{{\cos }^4}3x}} = {{6\sin 3x} \over {{{\cos }^3}3x}}\)

b. \(y' = \left({{\cos \sqrt {{x^2} + 1} } \over {\sqrt {{x^2} + 1} }}\right)' \)

\(= {{(cos\sqrt {{x^2} + 1} )'\sqrt {{x^2} + 1} - (\sqrt {{x^2} + 1} )'cos\sqrt {{x^2} + 1} } \over {{x^2} + 1}} \)

\(= {{ - sin\sqrt {{x^2} + 1} (\sqrt {{x^2} + 1} )'\sqrt {{x^2} + 1} - (\sqrt {{x^2} + 1} )'cos\sqrt {{x^2} + 1} } \over {{x^2} + 1}} \)

\(= {{ - sin\sqrt {{x^2} + 1}.{x \over {\sqrt {{x^2} + 1} }}.\sqrt {{x^2} + 1} - {x \over {\sqrt {{x^2} + 1} }}\cos \sqrt {{x^2} + 1} } \over {{x^2} + 1}} \)

\(=\frac{\frac{x}{\sqrt{x^2+1}}.\left ( -\sqrt{x^2+1}.sin\sqrt{x^2+1}-cos\sqrt{x^2+1} \right )}{x^2+1}\)

\(= {{ - x(\sqrt {{x^2} + 1} \sin \sqrt {{x^2} + 1} + \cos \sqrt {{x^2} + 1} )} \over {{{(\sqrt {{x^2} + 1} )}^3}}}\)

c. \(y '= \left((2 - {x^2})cos\,x + 2x.sin\,x\right)'\)     

\(= (2 – x^2)’cos \,x + (2 – x^2)(cos\,x)’ + (2x)’sin\,x + 2x(sin\, x)’\)

\(= - 2x cos\,x – (2 – x^2)sin\, x + 2sin \,x + 2x\,cos\,x = x^2sin\,x\)

d) \(y = {{\sin \,x - x.cos\,x} \over {\cos \,x + x.\sin \,x}}\)

Đặt \(u = \sin \,x - x\cos \,x; v=\cos \,x + x{\mathop{\rm sin\,x}\nolimits}\)

Ta có: 

\(u' = \cos\, x - (cos\,x - x\,sin\,x) = x\sin \,x \)

\(v' = - \sin \,x + (\sin \,x + x\cos \,x) = x\cos\, x \)

\(\Rightarrow y' = {{x{\mathop{\rm sinx}\nolimits} (cosx + xsinx) - x\cos x(\sin x - x\cos x)} \over {{{(cosx + x\sin x)}^2}}} \)

\(= \frac{x.sin\,x\,cos\,x+x^2\,sin^2\,x-x.cos\,x\,sin\,x+x^2\,cos^2\,x}{(cos\,x+x\,sin\,x)^2} \)

\(= {{{x^2}.(sin^2 x+cos^2 x)} \over {{{(cosx + xsinx)}^2}}} = {{{x^2}} \over {{{(cosx + xsinx)}^2}}}\)