Question

$if \: y = \sqrt{1 - 2x { \:}^{2} } \: find \: \frac{dy}{dx}$
​

Answer 1

Answer:

$y = \sqrt{1 - 2x {}^{2} }$

$\frac{dy}{dx} = \frac{d \sqrt{1 - 2x {}^{2} } }{dx}$

$\frac{d}{dx} \sqrt{1} = \\ \frac{d}{dx} 1 = 0$

$\frac{d}{dx} \sqrt{2x {}^{2} } = \\ \frac{d}{dx} \sqrt{2} \: x$

$\sqrt{2} \frac{d}{dx} x + x \frac{d}{dx} \sqrt{2}$

$\sqrt{2} \times 1 + x \times 0 \\ \sqrt{2 } + 0 \\ \sqrt{2}$

$\sqrt{2}$

is the answer ☺️