Tìm đạo hàm của các hàm số sau

$$\begin{gathered} a, y=5Sinx-3\cos x \\ b,y=\frac{\sin x+\cos x}{\sin x-cos} \\ c,y=xcotx \\ d,y=\frac{\sin x}{x}+\frac{x}{\sin x} \\ e, y=\sqrt{1+2\tan x} \\ f,y=\sin \sqrt{a+x^2} \\ Giải \\ a, y\text{'}=5\cos x+3\sin x \\ b,y\text{'}=\frac{\left(sinx+cosx \right)\text{'}\left(sinx-\cos x\right)-\left(\sin x+\cos x\right)\left(\sin x-\cos x\right)\text{'}}{{\left(\sin x-\cos x\right)}^2} \\ =\frac{\left(\cos x-sinx\right)\left(\sin x-\cos x\right)-\left(\sin x+\cos x\right)\left(cosx+\sin x\right)}{{\left(\sin x-\cos x\right)}^2} \\ =\frac{-{\left(\cos x-\sin x\right)}^2-{\left(\sin x-\cos x\right)}^2}{{\left(\sin x-\cos x\right)}^2}=\frac{-2}{\left(\sin x-\mathrm{co}sx \right)} \\ c,y\text{'}=cotx +x.\frac{\left(-1\right)}{{\sin }^{2}x}=cotx-\frac{x}{{\sin }^{2}x} \\ d, y\text{'}=\frac{x\cos x -\sin x }{x^2}+\frac{\sin x-x\cos x }{{\sin }^{2}x}=\left(x\cos x -sinx\right).\left(\frac{1}{x^2}-\frac{1}{{\sin }^{2}x}\right) \\ e, y\text{'}=\frac{{\left(1+2\tan x \right)}^2}{2\sqrt{1+2tanx }}=\frac{1}{ co{s}^{2}x\sqrt{1+2\tan }x} \\ f.y\text{'}=\frac{x}{\sqrt{1+x^2}}.\cos \sqrt{1+x^2} \end{gathered}$$