Question

How do you find the derivative of ${(1 + x)^{\dfrac{1}{x}}}$?

Answer 1

We solve for the derivative of the given term by assuming the term as a variable and then take log function on both sides of the equation. Use the property of logarithm and open the log on the right hand side of the equation. Differentiate both sides of the equation and use the product rule of differentiation on right hand side of the equation.

• If m and n are two integers then $m\log (n) = \log ({n^m})$
• Differentiation of${x^n}$with respect to x is given by $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
• Product rule of differentiation: $\dfrac{d}{{dx}}(m \times n) = m \times \dfrac{{dn}}{{dx}} + n \times \dfrac{{dm}}{{dx}}$

Complete step-by-step answer:
We are given the function ${(1 + x)^{\dfrac{1}{x}}}$
Let us assume $y = {(1 + x)^{\dfrac{1}{x}}}$
Take log function on both sides of the equation
$\Rightarrow \log y = \log \left[ {{{(1 + x)}^{\dfrac{1}{x}}}} \right]$
Since we know the rule of logarithm that $m\log (n) = \log ({n^m})$
$\Rightarrow \log y = \dfrac{1}{x}\log (1 + x)$
Now differentiate both sides of the equation with respect to x
$\Rightarrow \dfrac{d}{{dx}}\left( {\log y} \right) = \dfrac{d}{{dx}}\left( {\dfrac{1}{x}\log (1 + x)} \right)$
Use the formula of differentiation of log of a function i.e. $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$in left hand side of the equation and apply product rule of differentiation in right hand side of the equation.
$\Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = \log (1 + x)\dfrac{d}{{dx}}\left( {\dfrac{1}{x}} \right) + \dfrac{1}{x}\dfrac{d}{{dx}}\left( {\log (1 + x)} \right)$
Use formula of differentiation of log of a function i.e. $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$in right hand side of the equation
$\Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = \log (1 + x) \times \dfrac{d}{{dx}}{x^{ - 1}} + \dfrac{1}{x} \times \dfrac{1}{x}$
Use general formula of differentiation i.e. $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ in right hand side of the equation
$\Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = \log (1 + x) \times ( - 1){x^{ - 1 - 1}} + \dfrac{1}{{{x^2}}}$
$\Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = - \log (1 + x){x^{ - 2}} + \dfrac{1}{{{x^2}}}$
$\Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = - \log (1 + x)\dfrac{1}{{{x^2}}} + \dfrac{1}{{{x^2}}}$
Take term common in right hand side of the equation
$\Rightarrow \dfrac{1}{y} \times \dfrac{{dy}}{{dx}} = \dfrac{1}{{{x^2}}}\left( {1 - \log (1 + x)} \right)$
Substitute the value of y in left hand side of the equation
$\Rightarrow \dfrac{1}{{{{(1 + x)}^{\dfrac{1}{x}}}}} \times \dfrac{{dy}}{{dx}} = \dfrac{1}{{{x^2}}}\left( {1 - \log (1 + x)} \right)$
Cross multiply the value from denominator in left hand side of the equation to right hand side of the equation
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{x^2}}}\left( {1 - \log (1 + x)} \right){(1 + x)^{\dfrac{1}{x}}}$
Write the right hand side of the equation in proper manner
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{{(1 + x)}^{\dfrac{1}{x}}}}}{{{x^2}}}\left( {1 - \log (1 + x)} \right)$

$\therefore$The derivative of ${(1 + x)^{\dfrac{1}{x}}}$is $\dfrac{{{{(1 + x)}^{\dfrac{1}{x}}}}}{{{x^2}}}\left( {1 - \log (1 + x)} \right)$

Note:
Many students make the mistake of assuming the power given in terms of x as a power and directly apply the differentiation method which is wrong as that formula is for the function or value having constant value in power, not a function in power. Keep in mind we always use log method like done in the above solution when the power has a function.