# The curve $y=x^\frac{1}{5}$ has at

$\begin{array}{1 1} (A)\;\text{a vertical tangent (parallel to y-axis)}\\(B)\;\text{a horizontal tangent (parallel to x-axis)}\\(C)\;\text{an oblique tangent}\\(D)\;\text{no tangent}\end{array}$

Toolbox:
• Slope of a line is $ax+by+c=0$ is $- \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
• If $y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p =$ slope of the tangent to $y = f(x)$ at point $p$
Step 1
The equation of the curve is
$y=x^{\large\frac{1}{5}}$
differentiting w.r.t $x$ we get,
$\large\frac{dy}{dx} = \large\frac{1}{5}x^{\large\frac{-4}{5}}=\large\frac{1}{5x^{\large\frac{4}{5}}}$
The point given is $(0, 0)$
$\Rightarrow \large\frac{dy}{dx_{(0,0)}}=\large\frac{1}{0}$$=\infty$
The slope of the tangent is $\infty$
$\Rightarrow\:$ The tangent is parallel to $y$ - axis
Hence the correct option is A

edited Jan 12, 2014
when the slope of tangent is zero, is it not parallel to x-axis
Jitender,  slope of tangent in the above case is not zero, at  (0,0)  but it is 1/0,  which is infinity.  Therefore the tangent is parallel to y-axis.