Question

question
find $\frac{dy}{dx}$ if $y = sqrt{1 - 3x^{2}}$.
provide your answer below:
$\frac{dy}{dx}=square$

Explanation:

Step1: Rewrite the function

Rewrite $y = \sqrt{1 - 3x^{2}}$ as $y=(1 - 3x^{2})^{\frac{1}{2}}$.

Step2: Apply the chain - rule

The chain - rule states that if $y = f(g(x))$, then $\frac{dy}{dx}=f'(g(x))\cdot g'(x)$. Let $u = 1 - 3x^{2}$, so $y = u^{\frac{1}{2}}$. First, find $\frac{dy}{du}$ and $\frac{du}{dx}$.
$\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ (using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$ with $n=\frac{1}{2}$), and $\frac{du}{dx}=-6x$ (using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$ for $a=-3$ and $n = 2$, and $\frac{d}{dx}(c)=0$ for $c = 1$).

Step3: Calculate $\frac{dy}{dx}$

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{1}{2}(1 - 3x^{2})^{-\frac{1}{2}}\cdot(-6x)$.
Simplify the expression: $\frac{dy}{dx}=\frac{-3x}{\sqrt{1 - 3x^{2}}}$.

Answer:

$\frac{-3x}{\sqrt{1 - 3x^{2}}}$