# Calculus Derivatives - Finding the slope of a function

I am having trouble solving this question:

Consider the function $f(x)= 2x^{5/3} - 5x^{2/3}$

Determine the slope of the tangent at the point where the graph crosses the x-axis.

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We firstly determine the point where the graph crosses the x-axis. So we solve $f(x) = 0$ that is $$2x^{(2/3+3/3)}-5x^{(2/3)}=0\Rightarrow2x\cdot x^{(2/3)} - 5x^{(2/3)} = 0 \Leftrightarrow x^{(2/3)}(2x-5)=0$$ which gives $x = 5/2 \text{ or } x=0.$ Then we calculate the derivative of $y$ $$y'(x) = \frac{10}{3}x^{(2/3)} - \frac{10}{3}x^{(-1/3)}$$ which gives $$y'(x) = \frac{10}{3}\left(1-\frac{1}{x}\right)x^{(2/3)}$$ For $x=0$ the derivative does not exist so for $x = \frac{5}{2}$ we have that $$y'\left(\frac{5}{2}\right) = \frac{10}{3}\cdot\left(1-\frac{2}{5}\right)\cdot\left(\frac{5}{2}\right)^{(2/3)} =3.684$$.

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Solve f(x) = 0 ==> 2x*x^(2/3) - 5x^(2/3) = 0 ==> x = 0, 5/2. So y'(x) = 10/3x^(2/3) - 10/3x^(-1/3). We can't take x = 0 since y'(0) is not defined and the only slope is y'(5/2). y'(x) = 10/3(x - 1)/x^(2/3) . Evaluated at x = 5/2 gives y'(5/2) = (10/3*(5/2 - 1))/(5/2)^(2/3) = 5*(5/2)^(-2/3).

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Please use Latex in the answers. –  Jose Antonio Feb 27 '14 at 1:29