Bài 9.25 trang 97

Đề bài

Tính đạo hàm của các hàm số sau:

a) $y={\left(\frac{2x-1}{x+2}\right)}^5$

b) $y=\frac{2x}{x^2+1}$;

c) $y=e^x{\sin }^{2}x$;                                                       

d) $y=\log (x+\sqrt{x})$.

Lời giải chi tiết

a) $y^{\prime }=5{\left(\frac{2x-1}{x+2}\right)}^4{\left(\frac{2x-1}{x+2}\right)}^{,}=5{\left(\frac{2x-1}{x+2}\right)}^4.\frac{2\left(x+2\right)-\left(2x-1\right)}{{\left(x+2\right)}^2}=5{\left(\frac{2x-1}{x+2}\right)}^4.\frac{5}{{\left(x+2\right)}^2}=\frac{25{\left(2x-1\right)}^4}{{\left(x+2\right)}^6}$                                 

b) $y^{\prime }=\frac{{\left(2x\right)}^{\prime }\left(x^2+1\right)-2x{\left(x^2+1\right)}^{\prime }}{{\left(x^2+1\right)}^2}=\frac{2x^2+2-4x^2}{{\left(x^2+1\right)}^2}=\frac{-2x^2+2}{{\left(x^2+1\right)}^2}$;

c) $y^{\prime }={\left(e^x\right)}^{\prime }{\sin }^{2}x+e^x{\left({\sin }^{2}x\right)}^{\prime }=e^x{\sin }^{2}x+e^x.2\sin x.\cos x=e^x{\sin }^{2}x+e^x\sin 2x$;                                                     

d) $y^{\prime }={\left[\log \left(x+\sqrt{x}\right)\right]}^{,}=\frac{{\left(x+\sqrt{x}\right)}^{\prime }}{\left(x+\sqrt{x}\right)\ln 10}=\frac{1+\frac{1}{2\sqrt{x}}}{\left(x+\sqrt{x}\right)\ln 10}=\frac{2\sqrt{x}+1}{2\sqrt{x}\left(x+\sqrt{x}\right)\ln 10}$