Giải câu 3 bài 3: Đạo hàm của hàm số lượng giác

a) \(y = 5sinx -3cosx\)

\(y'=5cosx-3(-sinx)=5cosx+3sinx\)

b) \( y=\frac{sinx+cosx}{sinx-cosx}\)

\(y'={{(sinx+cos x)'.(sin x- cos x)-(sin x+cos x)(sin x-cos x)'}\over{(sin x-cos x)^{2}}}\)

\(= {{(cos x-sin x)(sin x -cos x)-(sin x+ cos x)(cosx+sinx)}\over{(sin x-cosx )^{2}}}\)

\( ={{-2}\over{(sin x-cos x)^{2}}}\)

c) \(y = x cotx\)

\(y' = cotx +x. \left ( -\frac{1}{sin^{2}x} \right )= cotx - \frac{x}{sin^{2}x}\).

d) \(y =  \frac{sinx}{x}+ \frac{x}{sinx}\)

\( y'=\frac{(sin x)'.x-sin x.(x)'}{x^{2}}+\frac{(x)'.sin x-x(sin x)'}{sin^{2}x}\)

\(= \frac{x.cosx-sinx}{x^{2}}+\frac{sin x-x.cosx}{sin^{2}x}\)

\(= \frac{x.cosx-sinx}{x^{2}}-\frac{x.cosx-sin x}{sin^{2}x}\)

\( = (x. cosx -sinx) \left ( \frac{1}{x^{2}}-\frac{1}{sin^{2}x} \right )\).

e) \(y = \sqrt{(1 +2tan x)}\)

\( y'=\frac{(1+2tanx)'}{2\sqrt{1+2tanx}}\)

\(= \frac{\frac{2}{cos^{2}x}}{2\sqrt{1+2tanx}}\)

\( =\frac{1}{cos^{2}x\sqrt{1+2tanx}}\).

f) \(y = sin\sqrt{(1 +x^2)}\)

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

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

\(=  \frac{2x}{2\sqrt{1+x^{2}}}cos\sqrt{(1+x^2)} \)

\(= \frac{x}{\sqrt{1+x^{2}}}cos\sqrt{(1+x^2)}\)