Cho f(x) = xsinx và $g\left(x\right)=\frac{\cos x}{x}$ . Giá trị $\frac{f^{\text{'}}\left(1\right)}{g^{\text{'}}\left(1\right)}$  là

A. −1.

B. sin1 + cos1.

C. 1.

D. −sin1 − cos1.

Đáp án đúng là: A

Có f'(x) = (xsinx)' = sinx + xcosx, suy ra f'(1) = sin1 + cos1

$g^{\text{'}}\left(x\right)={\left(\frac{\cos x}{x}\right)}^{\text{'}}$$=\frac{{\left(\cos x\right)}^{\text{'}}x-\cos x\cdot {\left(x\right)}^{\text{'}}}{x^2}$$=\frac{-x\mathrm{s}\text{inx}-\cos x}{x^2}$

suy ra $g^{\text{'}}\left(1\right)=\frac{-\mathrm{s}\text{in1}-\cos 1}{1^2}=-\mathrm{s}\text{in1}-\cos 1$

Do đó $\frac{f^{\text{'}}\left(1\right)}{g^{\text{'}}\left(1\right)}=\frac{\mathrm{s}\text{in1+}\cos 1}{-\mathrm{s}\text{in1}-\cos 1}=-1$

Vậy $\frac{f^{\text{'}}\left(1\right)}{g^{\text{'}}\left(1\right)}=-1$