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# Help with differentiation: Product rule.

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One of the question:

Consider y = Root(x) * (3-x)2

The dy/dx: [(3-x)(3-5x)] / (2 * root(x))

Find the x-coordinates of all points on y = Root(x) * (3-x) where the tangent is horizontal.

I thought the minimum point but im not entirely sure.

Guest Jul 29, 2017
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#1
+5552
+3

Your derivative is correct...here are the steps...

$$\begin{array}\ y&=&x^{\frac12} \cdot(3-x)^2 \\ \frac{dy}{dx}&=&(\frac{d}{dx} x^{\frac12})(3-x)^2+(\,\frac{d}{dx}(3-x)^2\,)(x^{\frac12}) \\ \frac{dy}{dx}&=&(\frac12)(x^{-\frac12})(3-x)^2+(2)(3-x)(\,\frac{d}{dx}(3-x)\,)(x^{\frac12}) \\ \frac{dy}{dx}&=&(\frac12)(x^{-\frac12})(3-x)^2+(2)(3-x)(-1)(x^{\frac12}) \\ \frac{dy}{dx}&=&(3-x)(\quad(\frac12)(x^{-\frac12})(3-x)+(2)(-1)(x^{\frac12})\quad) \\ \frac{dy}{dx}&=&(3-x)(\frac12)(\quad(x^{-\frac12})(3-x)-4x^{\frac12}\quad) \\ \frac{dy}{dx}&=&(3-x)(\frac12)(x^{-\frac12})(\quad(3-x)-4x\quad) \\ \frac{dy}{dx}&=&(3-x)(\frac12)(x^{-\frac12})(3-5x) \end{array}$$

The above equation tells us what the slope of this curve is at any  x  .

Where the tangent line is horizontal, the  slope  of the curve  =  0 .

Where the tangent line is horizontal, the  dy/dx  of the curve  =  0  .

We want to know what values for  x  cause  dy/dx  to be  0  .

$$0\,=\,(3-x)(\frac12)(x^{-\frac12})(3-5x)$$

Set the first and fourth factors equal to zero and solve for  x .

0  =  3 - x                   0    =   3 - 5x

x  =  3                        5x  =   3

x    =   3/5

You're right that the minumum is a point where the tangent line is horizontal...but the maximum is also a point where the tangent line is horizontal.

hectictar  Jul 29, 2017
#2
+91229
+3

y = Root(x) * (3-x)2

Find the x-coordinates of all points on y = Root(x) * (3-x)2  where the tangent is horizontal.

The first derivative GIVES the grandient of the tangent at any specific or general point. It is very important that you understand and remember this!

If the tangent is horizontal then the gradient of the tangent is 0.

So yes, you are being asked where dy/dx=0

$$y = \sqrt x * (3-x)^2\\ y = x ^{0.5}* (3-x)^2\\ \frac{dy}{dx}=0.5x^{-0.5}(3-x)^2\quad+\quad x^{0.5}*2(3-x)*-1\\ \frac{dy}{dx}=\frac{(3-x)^2}{2x^{0.5}}\quad-\quad 2(3-x)x^{0.5}\\ \frac{dy}{dx}=\frac{(3-x)^2}{2x^{0.5}}\quad-\quad 2(3-x)x^{0.5}*\frac{2x^{0.5}}{2x^{0.5}}\\ \frac{dy}{dx}=\frac{(3-x)^2}{2x^{0.5}}\quad-\quad \frac{4(3-x)x}{2x^{0.5}}\\ \frac{dy}{dx}=\frac{(3-x)^2-4(3-x)x}{2x^{0.5}}\quad\\ \frac{dy}{dx}=\frac{(3-x)[(3-x)-4x]}{2x^{0.5}}\quad\\ \frac{dy}{dx}=\frac{(3-x)(3-5x)}{2x^{0.5}}\quad\\$$

This will equal zero when  3-x=0   and when  3-5x=0

That is when x=3 and when x= 0.6

Substitute these points back inot the original equation to find the vaues of y.

$$When\;\; x=3\\ y=\sqrt3(0)=0\\ When\;\; x=0.6\\ y=\sqrt{0.6}(3-0.6)^2\\ y=\frac{\sqrt{3}}{\sqrt5}\frac{12^2}{5^2}\\ y=\frac{144\sqrt{3}}{25\sqrt5}\cdot\frac{\sqrt5}{\sqrt5}\\ y=\frac{144\sqrt{15}}{125}\\~\\ \text{The tangent to the curve will be horizontal}\\ \text{ at the points} (3,\frac{3}{5})\;\;and\;\;(\frac{3}{5},\frac{144\sqrt{15}}{125})$$

You should check my working. I have not done so there could be errors.

Melody  Jul 29, 2017

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