Question

tìm đạo hàm của mỗi hàm số sau :

a) y=\(\dfrac{1+x}{\sqrt{1-x}}\) ; b)y=\(\dfrac{x}{\sqrt{a^2-x^2}}\)

Answer 1

\(y=\dfrac{\left(x+1\right)}{\sqrt{1-x}}\)

\(y^2=\dfrac{\left(x+1\right)^2}{1-x}\)

\(y'=\dfrac{2\left(x+1\right)\left(1-x\right)+\left(x+1\right)^2}{2.\left(1-x\right)^2.\dfrac{\left(x+1\right)}{\sqrt{1-x}}}\)

Answer 2

cau a: \(y'=\dfrac{\sqrt{1-x}-\left(1+x\right)\left(\sqrt{1-x}\right)^'}{1-x}=\dfrac{\sqrt{1-x}-\left(1+x\right)\left(\dfrac{-1}{2\sqrt{1-x}}\right)}{1-x}=\dfrac{3-x}{2\left(\sqrt{1-x}\right)^2}\)câu b: \(y'=\dfrac{\sqrt{a^2-x^2}-x\left(\dfrac{-2x}{2\sqrt{a^2-x^2}}\right)}{a^2-x^2}=\dfrac{a^2}{\left(\sqrt{a^2-x^2}\right)^3}\)

Answer 3

câu b

\(y^2=\dfrac{x^2}{a^2-x^2}=\dfrac{a^2-\left(a^2-x^2\right)}{a^2-x^2}=\dfrac{a^2}{a^2-x^2}-1\)

\(y^2=\dfrac{a}{2}\left[\dfrac{1}{a-x}+\dfrac{1}{a+x}\right]-1\)

\(y^2=\dfrac{a}{2}\left[\dfrac{1}{a-x}+\dfrac{1}{a+x}\right]\)

\(VP'=\dfrac{a}{2}\left[\dfrac{1}{\left(a-x\right)^2}-\dfrac{1}{\left(a+x\right)^2}\right]=\dfrac{2a^2x}{\left(a^2-x^2\right)^2}\)

\(y'=\dfrac{4a^2x}{\dfrac{4.x}{\sqrt{a^2-x^2}}\left(a^2-x^2\right)^2}=\dfrac{a^2}{\sqrt{\left(a^2-x^2\right)^3}}\)

khác gì đâu

mình dùng lớp 8 phân tích nhỏ ra thôi :

cần mỗi cái này: \(y=ax^n\Rightarrow y'=anx^{n-1}\)

bắt hết các loại gió mùa

@nguyen ngoc song thuy