Question

find $f(x)$ given $f(x)=arcsin(3x)sqrt{1 - 9x^{2}}$. $f(x)=$ symbolic expression

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u=\arcsin(3x)$ and $v = \sqrt{1 - 9x^{2}}=(1 - 9x^{2})^{\frac{1}{2}}$.

Step2: Find $u'$

The derivative of $\arcsin(u)$ with respect to $x$ is $\frac{u'}{\sqrt{1 - u^{2}}}$. For $u = 3x$, $u'=3$. So, $u'=\frac{3}{\sqrt{1-(3x)^{2}}}=\frac{3}{\sqrt{1 - 9x^{2}}}$.

Step3: Find $v'$

Using the chain - rule, if $y=(1 - 9x^{2})^{\frac{1}{2}}$, let $t = 1 - 9x^{2}$, then $y=t^{\frac{1}{2}}$. The derivative of $y$ with respect to $t$ is $\frac{1}{2}t^{-\frac{1}{2}}$, and the derivative of $t$ with respect to $x$ is $-18x$. So, $v'=\frac{1}{2}(1 - 9x^{2})^{-\frac{1}{2}}\cdot(-18x)=\frac{-9x}{\sqrt{1 - 9x^{2}}}$.

Step4: Calculate $f'(x)$

$f'(x)=u'v+uv'$. Substitute $u, v, u', v'$ into the product - rule formula:
\[

$$\begin{align*} f'(x)&=\frac{3}{\sqrt{1 - 9x^{2}}}\cdot\sqrt{1 - 9x^{2}}+\arcsin(3x)\cdot\frac{-9x}{\sqrt{1 - 9x^{2}}}\\ &=3-\frac{9x\arcsin(3x)}{\sqrt{1 - 9x^{2}}} \end{align*}$$

\]

Answer:

$3-\frac{9x\arcsin(3x)}{\sqrt{1 - 9x^{2}}}$