tìm đạo hàm của các hàm số sau: a) y= 3 phần (x ² + x +1) ³ b) y=xcănx c) y = căn x-1+cănx+2 d) y= x √1+x ² 21/08/2021 Bởi Julia tìm đạo hàm của các hàm số sau: a) y= 3 phần (x ² + x +1) ³ b) y=xcănx c) y = căn x-1+cănx+2 d) y= x √1+x ²
a, $y’=\dfrac{-3[(x^2+x+1)^3]’}{(x^2+x+1)^6}$ $=\dfrac{-3.3(x^2+x+1)^2.(x^2+x+1)’}{(x^2+x+1)^6}$ $=\dfrac{-9.(2x+1)}{(x^2+x+1)^4}$ b, $y’=x’\sqrt{x}+x.(\sqrt{x})’=\sqrt{x}+x.\dfrac{1}{2\sqrt{x}}=\sqrt{x}+\dfrac{1}{2}.\sqrt{x}=\dfrac{3\sqrt{x}}{2}$ c, $y’=(\sqrt{x-1})’+(\sqrt{x+2})’=\dfrac{1}{2\sqrt{x-1}}+\dfrac{1}{2\sqrt{x+2}}$ d, $y’=x’\sqrt{1+x^2}+x.(\sqrt{1+x^2})’=\sqrt{1+x^2}+\dfrac{x.(1+x^2)’}{2\sqrt{1+x^2}}=\sqrt{1+x^2}+\dfrac{x^2}{\sqrt{1+x^2}}=\dfrac{2x^2+1}{\sqrt{1+x^2}}$ Bình luận
a,
$y’=\dfrac{-3[(x^2+x+1)^3]’}{(x^2+x+1)^6}$
$=\dfrac{-3.3(x^2+x+1)^2.(x^2+x+1)’}{(x^2+x+1)^6}$
$=\dfrac{-9.(2x+1)}{(x^2+x+1)^4}$
b,
$y’=x’\sqrt{x}+x.(\sqrt{x})’=\sqrt{x}+x.\dfrac{1}{2\sqrt{x}}=\sqrt{x}+\dfrac{1}{2}.\sqrt{x}=\dfrac{3\sqrt{x}}{2}$
c,
$y’=(\sqrt{x-1})’+(\sqrt{x+2})’=\dfrac{1}{2\sqrt{x-1}}+\dfrac{1}{2\sqrt{x+2}}$
d,
$y’=x’\sqrt{1+x^2}+x.(\sqrt{1+x^2})’=\sqrt{1+x^2}+\dfrac{x.(1+x^2)’}{2\sqrt{1+x^2}}=\sqrt{1+x^2}+\dfrac{x^2}{\sqrt{1+x^2}}=\dfrac{2x^2+1}{\sqrt{1+x^2}}$