Question

How do you find the derivative of ${e^{\dfrac{1}{{2x}}}}$ ?

Answer 1

In this question, we are given a line whose equation is ${\text{y - 5 = 3(x - 2)}}$ and we have been asked to find out or change the equation into intercept form. We can form it to its slope intercept form of the given equation ${\text{y - 5 = 3(x - 2)}}$ by substituting the values in the given formula.

Formulas used:
For any equation ${\text{Ax}} + {\text{ By }} + \;{\text{C}}\; = \;0$ ,
Slope (m) = $\dfrac{{ - {\text{A}}}}{{\text{B}}}$
Slope intercepts form:
${\text{y = mx + b}}$

Complete step-by-step answer:
We have to find the derivative of ${e^{\dfrac{1}{{2x}}}}$ .
To derive $e$ , we use this formula $\dfrac{d}{{dx}}{e^x} = {e^x}$ .
Since, in the given question, there is a variable, it also should be derivative according to the rule or formula in derivative $\dfrac{d}{{dx}}{e^{2x}} = 2{e^{2x}}$ .
So now, derivation of ${e^{\dfrac{1}{{2x}}}}$ ,
Using the formula, we first derivate ${e^{\dfrac{1}{{2x}}}}$ and then, the variable $\dfrac{1}{{2{\text{x}}}}$ ,
$\dfrac{d}{{dx}}{e^{\dfrac{1}{{2x}}}} = {e^{\dfrac{1}{{2x}}}} \cdot \dfrac{d}{{dx}}\dfrac{1}{{2{\text{x}}}}$
Now, using the formula of derivation of $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right)$ , we get
$\dfrac{d}{{dx}}{e^{\dfrac{1}{{2x}}}} = {e^{\dfrac{1}{{2x}}}} \cdot \dfrac{{\dfrac{d}{{dx}}(1) \cdot (2{\text{x) - }}\dfrac{d}{{dx}}(2{\text{x)}} \cdot (1)}}{{{{\left( {2{\text{x}}} \right)}^2}}}$
Derivate and simplifying the term we get,
$\dfrac{d}{{dx}}{e^{\dfrac{1}{{2x}}}} = {\text{ }}{e^{\dfrac{1}{{2x}}}} \cdot \dfrac{{0 - 2(1)}}{{{\text{4}}{{\text{x}}^2}}}$
Since $2$ is multiple of $4$ , we can cancel it.
$\dfrac{d}{{dx}}{e^{\dfrac{1}{{2x}}}} = {\text{ }}{e^{\dfrac{1}{{2x}}}} \cdot \dfrac{{ - 2}}{{{\text{4}}{{\text{x}}^2}}}$
$\dfrac{d}{{dx}}{e^{\dfrac{1}{{2x}}}} = {\text{ }} - \dfrac{1}{{{\text{2}}{{\text{x}}^2}}}{e^{\dfrac{1}{{2x}}}}$
Therefore, derivative of ${e^{\dfrac{1}{{2x}}}}$is${\text{ }} - \dfrac{1}{{{\text{2}}{{\text{x}}^2}}}{e^{\dfrac{1}{{2x}}}}$.

Note:
Alternative method:
We can solve this question using the logarithm method. First, let us assume the given question as some variable.
Let ${e^{\dfrac{1}{{2x}}}}$ be${\text{y}}$,
${\text{y = }}{e^{\dfrac{1}{{2x}}}}$
Now, we will solve this by using the logarithm method.
Taking $\log$ on both the sides, we get
${\text{y = }}{e^{\dfrac{1}{{2x}}}}$ Becomes ${\text{log y = log }}{e^{\dfrac{1}{{2x}}}}$ and now,
${\text{log y = log }}{e^{\dfrac{1}{{2x}}}}$
Using the property of logarithm, the power of $e$ will be multiplied to the term,
${\text{log y = }}\dfrac{1}{{2{\text{x}}}}{\text{log }}e$
Now, taking derivative on both the sides, we get
Since we finding derivation of the given term with respect to ${\text{x}}$ , derivation of ${\text{y}}$ is $\dfrac{{dy}}{{dx}}$
Using the derivation of logarithm formula, we get
$\dfrac{1}{{\text{y}}}\dfrac{{dy}}{{dx}}{\text{ = }}\left( {\dfrac{d}{{dx}}{\text{ log }}e} \right)\left( {\dfrac{d}{{dx}}\dfrac{1}{{2{\text{x}}}}} \right)$
We know that $\left( {\dfrac{d}{{dx}}{\text{ log }}e} \right) = 1$and $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{\dfrac{d}{{dx}}u.v{\text{ - }}\dfrac{d}{{dx}}v.u}}{{{v^2}}}$we get,
$\dfrac{1}{{\text{y}}}\dfrac{{dy}}{{dx}}{\text{ = }}1.\left( {\dfrac{{\dfrac{d}{{dx}}(1) \cdot (2{\text{x) - }}\dfrac{d}{{dx}}(2{\text{x)}} \cdot (1)}}{{{{\left( {2{\text{x}}} \right)}^2}}}} \right)$
Simplifying the term,
$\dfrac{1}{{\text{y}}}\dfrac{{dy}}{{dx}}{\text{ = }}\dfrac{{0 - 2(1)}}{{{\text{4}}{{\text{x}}^2}}}$
$\dfrac{1}{{\text{y}}}\dfrac{{dy}}{{dx}}{\text{ = }}\dfrac{{ - 2}}{{{\text{4}}{{\text{x}}^2}}}$
$\dfrac{1}{{\text{y}}}\dfrac{{dy}}{{dx}}{\text{ = }}\dfrac{{ - 1}}{{{\text{2}}{{\text{x}}^2}}}$
Now transferring ${\text{y}}$ to the other side,
$\dfrac{{dy}}{{dx}}{\text{ = (y)}}\dfrac{{ - 1}}{{{\text{2}}{{\text{x}}^2}}}$
We know that ${\text{y = }}{e^{\dfrac{1}{{2x}}}}$, substituting it, it becomes
$\dfrac{{dy}}{{dx}}{\text{ = }}\dfrac{{ - 1}}{{{\text{2}}{{\text{x}}^2}}}.{e^{\dfrac{1}{{2x}}}}$is the required answer.