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Bài 17. Tính đạo hàm của các hàm số sau
a) \(y = {1 \over {{{\cos }^2}3x}}\)
b) \(y = {{\cos \sqrt {{x^2} + 1} } \over {\sqrt {{x^2} + 1} }}\)
c) \(y = (2 – {x^2})cosx + 2x.sinx\)
d) \(y = {{\sin x – x.cosx} \over {\cos x + x.\sin x}}\)
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)
\(\eqalign{
& y’ = \left({{\cos \sqrt {{x^2} + 1} } \over {\sqrt {{x^2} + 1} }}\right)’ \cr
& = {{(cos\sqrt {{x^2} + 1} )’\sqrt {{x^2} + 1} – (\sqrt {{x^2} + 1} )’cos\sqrt {{x^2} + 1} } \over {{x^2} + 1}} \cr
& = {{ – sin\sqrt {{x^2} + 1} (\sqrt {{x^2} + 1} )’\sqrt {{x^2} + 1} – (\sqrt {{x^2} + 1} )’cos\sqrt {{x^2} + 1} } \over {{x^2} + 1}} \cr
& = {{ – sin\sqrt {{x^2} + 1}.{x \over {\sqrt {{x^2} + 1} }}.\sqrt {{x^2} + 1} – {x \over {\sqrt {{x^2} + 1} }}\cos \sqrt {{x^2} + 1} } \over {{x^2} + 1}} \cr
& = {{ – x(\sqrt {{x^2} + 1} \sin \sqrt {{x^2} + 1} + \cos \sqrt {{x^2} + 1} )} \over {{{(\sqrt {{x^2} + 1} )}^3}}} \cr} \)
c)
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\(y ‘= \left((2 – {x^2})cosx + 2x.sinx\right)’\)
\(y’ = (2 – x^2)’cos x + (2 – x^2)(cosx)’ + (2x)’sinx + 2x(sin x)’\)
\(= – 2x cosx – (2 – x^2)sin x + 2sin x + 2xcosx = x^2sinx\)
d) \(y = {{\sin x – x.cosx} \over {\cos x + x.\sin x}}\)
\(\left\{ \matrix{
u = \sin x – x\cos x \Rightarrow u’ = \cos x – (cosx – xsinx) = x\sin x \hfill \cr
v = \cos x + x{\mathop{\rm sinx}\nolimits} \Rightarrow v’ = – \sin x + (\sin x + x\cos x) = x\cos x \hfill \cr} \right.\)
Vậy:
\(\eqalign{
& y’ = {{x{\mathop{\rm sinx}\nolimits} (cosx + xsinx) – x\cos x(\sin x – x\cos x)} \over {{{(cosx + x\sin x)}^2}}} \cr
& = {{{x^2}.(sin^2 x+cos^2 x)} \over {{{(cosx + xsinx)}^2}}} = {{{x^2}} \over {{{(cosx + xsinx)}^2}}} \cr} \)