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"Find the derivative of $y=x\sqrt{9-x}$."

So this is what I have and now I'm stuck.

\begin{align} y' &= x \frac{d}{dx}\left[(9-x)^{1/2}\right] + (9-x)^{1/2} \frac{d}{dx}(x)\\ &= x \left[\frac{1}{2}(9-x)^{-1/2}\right] + (9-x)^{1/2} (1) \end{align}

So I now that I need to multiply and simplify but I don't know where to start. Help!

This problem is actually part of a homework question where I have to analyze a graph and find critical points and min and max.

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    $\begingroup$ Note that $\frac{d}{dx}\sqrt{9-x}=\frac{-1}{2\sqrt{9-x}}.$ To simplify use that $\frac{-x}{2\sqrt{9-x}}+\sqrt{9-x}=\frac{-x+2(9-x)}{2\sqrt{9-x}}.$ $\endgroup$
    – mfl
    Jul 30 '14 at 17:54
  • $\begingroup$ OP wrote $-1/2$ for the exponent in the derivative of $\sqrt{9-x}$, which got removed by accident by an editor. Still forgot the minus sign outside the exponent. $\endgroup$ Jul 30 '14 at 18:12
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Just for fun, an alternate approach:

$$\begin{align} y&= x\sqrt{9-x}\\ y^2&=9x^2-x^3\\ 2yy'&=18x-3x^2\\ y'&=\frac{18x-3x^2}{2y}\\ y'&=\frac{18x-3x^2}{2x\sqrt{9-x}}\\ y'&=\frac{18-3x}{2\sqrt{9-x}} \end{align}$$

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Set $$f(x)= x, \quad g(z) = \sqrt{z}\quad \text{ and }\quad h(x)= 9-x,$$ then $$y = f(x)g(h(x)).$$ You should be able to compute $$f'(x)= 1, \quad g'(z) = \frac{1}{2\sqrt{z}}\quad \text{ and }\quad h'(x)=-1.$$ Now using the multiplication rule and the chain rule for derivatives, we know that $$y' = g(h(x))f'(x)+f(x)(g(h(x)))' = g(h(x))f'(x)+f(x)(g'(h(x))h'(x)),$$ replacing by the expressions above, we get $$y' = \sqrt{9-x}\cdot 1+x\left(\frac{1}{2\sqrt{9-x}}\cdot (-1)\right)= \sqrt{9-x}-\frac{x}{2\sqrt{9-x}}$$

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Also, for fun, a different approach using logarithms. It might seem it makes things harder, but it actually gets you to look at similar problems in a different way. If $f(x) = x \sqrt{9-x}$, then define $Lf(x) = \log f(x)$. You get $$ Lf(x) = \log x + \log \sqrt{9-x} = \log x +\frac{1}{2}\log (9-x)\\ \frac{d L f(x)}{dx} = \frac{f'(x)}{f(x)} = \frac{1}{x}-\frac{1}{2(9-x)} = \frac{3(6-x)}{2x(9-x)} $$ Hence, $$ f'(x) = \frac{3f(x)(6-x)}{2x(9-x)} = \frac{3x \sqrt{9-x}(6-x)}{2x(9-x)} = \frac{3 \sqrt{9-x}(6-x)}{2(9-x)} $$

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$$y=x\sqrt{9-x}$$ $$y'=x'\sqrt{9-x}+x(\sqrt{9-x})'=\sqrt{9-x}+x\frac{1}{2\sqrt{9-x}}(9-x)'=$$ $$=\sqrt{9-x}+x\frac{1}{2\sqrt{9-x}}(-1)=\sqrt{9-x}+\frac{-x}{2\sqrt{9-x}}=$$ $$=\frac{2(9-x)-x}{2\sqrt{9-x}}=\frac{18-3x}{2\sqrt{9-x}}$$

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An inspection of given expression shows x = 0 and x = 9 are roots.

When sqrt appears, it is often advantageous to remove it by squaring.

$ y^2 = 9 x^2-x^3, 2\ y\ y' = 18\,x -3x^2, 2 \,(y \,y'' + y^{'2})= 18-6 x $

@ x = 6 maximum is found as second derivative is < 0.

From the second relation the derivative is

$$ y'= \dfrac{3(6-x)}{2\sqrt{9-x}}$$

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Let $$ y=(-u+9)(u^{\frac{1}{2}})$$ Then, $$y=(-u^{3/2}+9u^{1/2})$$

And the derivate is, $$y'= -{\frac{3}{2}}(u^{1/2})+{\frac{9}{2}}(u^{-1/2})$$

As $u= 9-x$ Then,

$$y'= -{\frac{3}{2}}[(9-x)^{\frac{1}{2}}]+{\frac{9}{2}}[(9-x)^{\frac{-1}{2}}]$$

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