Calculus Derivatives  Finding the slope of a function
We firstly determine the point where the graph crosses the xaxis. So we solve $f(x) = 0$ that is $$2x^{(2/3+3/3)}5x^{(2/3)}=0\Rightarrow2x\cdot x^{(2/3)}  5x^{(2/3)} = 0 \Leftrightarrow x^{(2/3)}(2x5)=0$$ 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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user481710
Updated on August 01, 2022Comments

user481710 over 1 year
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 xaxis.

Jose Antonio over 9 yearsPlease use Latex in the answers.