Đáp án:
`a,y' =\frac{3x³ +9x}{\sqrt{x²+4}}`
`b, y' = 12sin^3 3x.cos3x`
Giải thích các bước giải:
`a,y= (x²+1) \sqrt{x²+4}`
`=> y' =( x²+1)' . \sqrt{x²+4} + (\sqrt{x²+4})' .(x²+1)`
`=>y' =2x .\sqrt{x²+4} + \frac{2x}{2\sqrt{x²+4}}.(x²+1)`
`=>y' = 2x. \sqrt{x²+4} +\frac{x³+x}{\sqrt{x²+4}}`
`=> y' =\frac{2x(x²+4)+x³+x}{\sqrt{x²+4}}`
`=> y' =\frac{2x³ +8x+x³ +x}{\sqrt{x²+4}}`
`=> y' =\frac{3x³ +9x}{\sqrt{x²+4}}`
___________________________
`b, y=sin^4 3x`
`=>y' =4sin^3 3x. (sin3x)'`
`=> y' = 4sin^3 3x .cos 3x.3`
`=> y' = 12sin^3 3x.cos3x`
