Tính đạo hàm ( x + $\sqrt[]{x^2+1}$ )$^{3}$ 27/08/2021 Bởi Gabriella Tính đạo hàm ( x + $\sqrt[]{x^2+1}$ )$^{3}$
$f(x)=(x+\sqrt{x^2+1})^3$ $f'(x)=3(x+\sqrt{x^2+1})^2.(x+\sqrt{x^2+1})’$ $=3(x+\sqrt{x^2+1})^2.\Big(1+\dfrac{2x}{2\sqrt{x^2+1}}\Big)$ $=3(x+\sqrt{x^2+1})^2.\dfrac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}$ $=\dfrac{3(x+\sqrt{x^2+1})^3}{\sqrt{x^2+1}}$ Bình luận
Đáp án: `y’ =\frac{3.(x+\sqrt{x²+1})^3}{\sqrt{x²+1}}` Giải thích các bước giải: `y=(x+\sqrt{x²+1})^3` `=> y’ = 3. (x+\sqrt{x²+1})^2 .(x+\sqrt{x²+1})’` `=> y’ = 3. (x+\sqrt{x²+1})^2 . ( 1 +\frac{2x}{2\sqrt{x²+1}})` `=> y’ =3(x +\sqrt{x²+1})^2 . \frac{\sqrt{x²+1} + x}{\sqrt{x²+1})` `=> y’ =\frac{3.(x+\sqrt{x²+1})^3}{\sqrt{x²+1}}` Bình luận
$f(x)=(x+\sqrt{x^2+1})^3$
$f'(x)=3(x+\sqrt{x^2+1})^2.(x+\sqrt{x^2+1})’$
$=3(x+\sqrt{x^2+1})^2.\Big(1+\dfrac{2x}{2\sqrt{x^2+1}}\Big)$
$=3(x+\sqrt{x^2+1})^2.\dfrac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}$
$=\dfrac{3(x+\sqrt{x^2+1})^3}{\sqrt{x^2+1}}$
Đáp án: `y’ =\frac{3.(x+\sqrt{x²+1})^3}{\sqrt{x²+1}}`
Giải thích các bước giải:
`y=(x+\sqrt{x²+1})^3`
`=> y’ = 3. (x+\sqrt{x²+1})^2 .(x+\sqrt{x²+1})’`
`=> y’ = 3. (x+\sqrt{x²+1})^2 . ( 1 +\frac{2x}{2\sqrt{x²+1}})`
`=> y’ =3(x +\sqrt{x²+1})^2 . \frac{\sqrt{x²+1} + x}{\sqrt{x²+1})`
`=> y’ =\frac{3.(x+\sqrt{x²+1})^3}{\sqrt{x²+1}}`