Question

Differentiate the following function w.r.t x.
$\sqrt {x + \dfrac{1}{x}}$

Answer 1

Use the various concept of differentiation as division rule as $\dfrac{{d\left( {\dfrac{a}{b}} \right)}}{{dx}} = \dfrac{{b\dfrac{{da}}{{dx}} - a\dfrac{{db}}{{dx}}}}{{{b^2}}}$ and also the differentiation of $\dfrac{{d\left( {\sqrt x } \right)}}{{dx}} = \dfrac{1}{{2\sqrt x }}$. And also remember the general differentiation formula as $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$. And hence put the values of all the above terms and simplify the equation and derivate the above.

Complete step by step answer:

As the given equation is as $\sqrt {x + \dfrac{1}{x}}$
Hence, simplify the above term as $\sqrt {\dfrac{{{x^2} + 1}}{x}}$
Now, apply the chain rule and we use $\dfrac{{d\left( {\sqrt x } \right)}}{{dx}} = \dfrac{1}{{2\sqrt x }}$ , we get,
\Rightarrow $$$$\dfrac{{d\left( {\sqrt {\dfrac{{{x^2} + 1}}{x}} } \right)}}{{dx}} = \dfrac{1}{{2\sqrt {\dfrac{{{x^2} + 1}}{x}} }}\dfrac{{d\left( {\dfrac{{{x^2} + 1}}{x}} \right)}}{{dx}}
Now using the division rule which is $\dfrac{{d\left( {\dfrac{a}{b}} \right)}}{{dx}} = \dfrac{{b\dfrac{{da}}{{dx}} - a\dfrac{{db}}{{dx}}}}{{{b^2}}}$, we get,
\Rightarrow $$$$\dfrac{{d\left( {\sqrt {\dfrac{{{x^2} + 1}}{x}} } \right)}}{{dx}} = \dfrac{1}{{2\sqrt {\dfrac{{{x^2} + 1}}{x}} }}\left( {\dfrac{{x\left( {2x} \right) - \left( {{x^2} + 1} \right)1}}{{{x^2}}}} \right)
On simplifying the above term as
$= \dfrac{{\sqrt x }}{{2\sqrt {{x^2} + 1} }}\left( {\dfrac{{\left( {2{x^2}} \right) - \left( {{x^2}} \right) - 1}}{{{x^2}}}} \right)$
On simplifying the numerator we get,
$= \dfrac{{\sqrt x }}{{2\sqrt {{x^2} + 1} }}\left( {\dfrac{{{x^2} - 1}}{{{x^2}}}} \right)$
On further simplification we get,
$= \dfrac{1}{{2\sqrt {{x^2} + 1} }}\left( {\dfrac{{{x^2} - 1}}{{{x^{\dfrac{3}{2}}}}}} \right)$
Hence, above is our required differentiation.

Note: Hence, use the above differentiation concept properly and solve all the given equation and simplify it and do the calculation error free. As using division rule, product rule, chain rule, and also remember various differentiation rule and hence, above required answer can be obtained.
Some basic differentiation formulas are,
$\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
$\dfrac{{d(ab)}}{{dx}} = a\dfrac{{db}}{{dx}} + b\dfrac{{da}}{{dx}}$
$\dfrac{{d\left( {\dfrac{a}{b}} \right)}}{{dx}} = \dfrac{{b\dfrac{{da}}{{dx}} - a\dfrac{{db}}{{dx}}}}{{{b^2}}}$