Name: Derivative of Functions in Parametric Form
Uploaded: 2019-07-19
Duration: 8 min 37 s
Description: Learn everything you'll need to know about derivatives of functions in parametric forms with this quick and easy lesson that'll cover the concept comprehensi...

Question

Find the exact coordinates of the stationary point on the curve

y = \frac{e^{2 x}}{x^{2}}

and determine its nature.

Accepted Answer

Given curve is $y = \frac{e^{2 x}}{x^{2}}$

We need to find the gradient of this curve i.e., $\frac{d y}{d x}$

$\frac{d y}{d x} = \frac{d}{d x} (\frac{e^{2 x}}{x^{2}})$

$= \frac{x^{2} \frac{d}{d x} (e^{2 x}) - e^{2 x} \frac{d}{d x} (x^{2})}{x^{4}}$ $[∵ \frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}]$

$= \frac{x^{2} \times e^{2 x} \times \frac{d}{d x} (2 x) - e^{2 x} \times 2 x}{x^{4}}$ $[∵ \frac{d}{d x} (e^{a x}) = e^{a x} \frac{d}{d x} (a x), \frac{d}{d x} (x^{n}) = n x^{n - 1}, n \in R]$

$= \frac{x^{2} e^{2 x} \times 2 - 2 x e^{2 x}}{x^{4}}$

$= \frac{2 x e^{2 x} (x - 1)}{x^{4}} . . . . . (i)$

Now, for the stationary points, we have

$\frac{d y}{d x} = 0$

$\Rightarrow x (x - 1) = 0$

$\Rightarrow x = 1$ [Since $y = \frac{e^{2 x}}{x^{2}}$ is not defined at $x = 0$ ]

The value of $y$ when $x = 1$ is

$y = \frac{e^{2 (1)}}{{(1)}^{2}} = e^{2}$

Hence, the exact coordinates of the stationary point on the given curve are $(1, e^{2})$ .

Now, let us find whether this is a point of minima or maxima

$\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x})$

$= \frac{d}{d x} (\frac{2 e^{2 x}}{x^{2}} - \frac{2 e^{2 x}}{x^{3}})$ [using $(i)$ ]

$= 2 \frac{d}{d x} (\frac{e^{2 x}}{x^{2}}) - 2 \frac{d}{d x} (\frac{e^{2 x}}{x^{3}})$

$= 2 [\frac{x^{2} \frac{d}{d x} (e^{2 x}) - e^{2 x} \frac{d}{d x} (x^{2})}{x^{4}}] - 2 [\frac{x^{3} \frac{d}{d x} (e^{2 x}) - e^{2 x} \frac{d}{d x} (x^{3})}{x^{6}}]$

$= 2 [\frac{x^{2} (e^{2 x}) \times 2 - e^{2 x} \times (2 x)}{x^{4}}] - 2 [\frac{x^{3} \times (e^{2 x}) \times 2 - e^{2 x} \times (3 x^{2})}{x^{6}}]$

$= \frac{4 x e^{2 x} (x - 1)}{x^{4}} - \frac{2 x^{2} e^{2 x} (2 x - 3)}{x^{6}}$

$= \frac{4 e^{2 x} (x - 1)}{x^{3}} - \frac{2 e^{2 x} (2 x - 3)}{x^{4}} . . . . (i i)$

Now, at $x = 1$ let us check the sign of $\frac{d^{2} y}{d x^{2}}$

$\frac{d^{2} y}{d x^{2}} = 0 - \frac{2 e^{2} (- 1)}{1} = 2 e^{2} > 0$ [using $(i i)$ ]

Thus, $(1, e^{2})$ is a point of local minima.

Find the exact coordinates of the stationary point on the curve $y = \frac{e^{2 x}}{x^{2}}$ and determine its nature.

Important Questions on Differentiation

Learn and Prepare for any exam you want !

Find the exact coordinates of the stationary point on the curve y=e2xx2 and determine its nature.

Important Questions on Differentiation

Learn and Prepare for any exam you want !

Find the exact coordinates of the stationary point on the curve $y = \frac{e^{2 x}}{x^{2}}$ and determine its nature.