So here is a problem that I am having a bit of trouble with.
Given $s(x) = \int_a^x \sqrt{1 + [f'(t)]^2} \ dt$, and $y = f(x)$, if $9x^2 = y(y - 3)^2$ find $\frac{ds}{dx}$, if it exists.
My attempt to this question was to start by applying FTC1 to $s(x)$, so \begin{align*} s'(x) &= \sqrt{1 + [f'(x)]^2} \end{align*}
With the function $s'(x)$ we can also say $\frac{ds}{dx}$. \begin{align*} \frac{ds}{dx} &= \sqrt{1 + [f'(x)]^2} \end{align*}
Then with $y = f(x)$, we can say that $\frac{dy}{dx} = f'(x)$. With our equation, we want to determine $\frac{dy}{dx}$. We can do so by implicit differentiation: \begin{align*} 9x^2 &= y(y - 3)^2 \\ 9x^2 &= y^3 - 6y^2 + 9y \\ 18x &= 3y^2 \cdot \frac{dy}{dx} - 12y \cdot \frac{dy}{dx} + 9 \cdot \frac{dy}{dx} \\ 18x &= \frac{dy}{dx}(3y^2-12y+9) \\ \frac{dy}{dx} &= \frac{18x}{3y^2 - 12y + 9} \\ \frac{dy}{dx} &= \frac{6x}{y^2 - 4y + 3} \end{align*}
So now I found $\frac{dy}{dx}$, we can square it so we can have $(\frac{dy}{dx})^2 = [f'(x)]^2$ \begin{align*} \left(\frac{dy}{dx}\right)^2 &= \frac{36x^2}{(y^2 - 4y + 3)^2} \end{align*}
Now putting what we know into $s'(x) = \frac{ds}{dx}$, \begin{align*} \frac{ds}{dx} &= \sqrt{1 + \frac{36x^2}{(y^2 - 4y + 3)^2}} \\ &= \sqrt{\frac{(y^2 - 4y + 3)^2 + 36x^2}{(y^2 - 4y + 3)^2}} \end{align*}
After this part, I am not even sure if I am doing the question right anymore. If possible would like some advice or clarification. Thanks
Given: $$9x^2=y(y-3)^2~~~(1)$$ We get $y'=\frac{6x}{y^2-4y+3}$. Next by Lebnitz, we get $$\frac{ds}{dx}=\sqrt{1+y'^2(x)}= \frac{\sqrt{4y(y-3)^2+(y-3)^2(y-1)^2}}{(y-3)(y-1)} =\frac{y(x)+1}{y(x)-1},$$ when $y\ne 1,3.$