Cho hàm số $y=\sqrt{1-x^{2}}$. Chứng minh rằng: $y \cdot y^{\prime}+x=0 ; \forall x \in(-1 ; 1)$.

Q: Cho hàm số $y=\sqrt{1-x^{2}}$. Chứng minh rằng: $y \cdot y^{\prime}+x=0 ; \forall x \in(-1 ; 1)$.

<p>- <span class="math-tex">$y^{\prime}=\frac{\left(1-x^{2}\right)^{\prime}}{2 \sqrt{1-x^{2}}}=\frac{-x}{\sqrt{1-x^{2}}} \Rightarrow y \cdot y^{\prime}=\sqrt{1-x^{2}} \cdot\left(\frac{-x}{\sqrt{1-x^{2}}}\right)=-x \Rightarrow y \cdot y^{\prime}+x=0$</span></p>

Question

Accepted Answer

- $y^{\prime}=\frac{\left(1-x^{2}\right)^{\prime}}{2 \sqrt{1-x^{2}}}=\frac{-x}{\sqrt{1-x^{2}}} \Rightarrow y \cdot y^{\prime}=\sqrt{1-x^{2}} \cdot\left(\frac{-x}{\sqrt{1-x^{2}}}\right)=-x \Rightarrow y \cdot y^{\prime}+x=0$

THPT Nguyễn Thị Minh Khai - Đề thi cuối kì 2 (CT) 18-19 - Q. 3 - Tp. Hồ Chí Minh - MĐ 6633