Question

How do you find the derivative of $\sqrt {\dfrac{1}{{{x^3}}}}$ ?

Answer 1

Hint : We can simplify the given problem to the simplest form then we apply the differentiation to it. We know the differentiation formula,
$\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$ .
We also know that if we have $\sqrt a$ then we can write it as ${a^{\dfrac{1}{2}}}$ . To solve this we need to know laws of indices.

Complete step-by-step answer :
Given, $\sqrt {\dfrac{1}{{{x^3}}}}$ .
We can write this as
$= \dfrac{{\sqrt 1 }}{{\sqrt {{x^3}} }}$
Square root of 1 is 1.
$= \dfrac{1}{{\sqrt {{x^3}} }}$
$= \dfrac{1}{{{{({x^3})}^{\dfrac{1}{2}}}}}$
Applying the law of brackets, ${\left( {{x^m}} \right)^n} = {x^{m \times n}}$ .
$= \dfrac{1}{{{x^{\dfrac{3}{2}}}}}$
If we shift the denominator term to the numerator we have,
$= {x^{ - \dfrac{3}{2}}}$ .
Thus we have,
$\Rightarrow \sqrt {\dfrac{1}{{{x^3}}}} = {x^{ - \dfrac{3}{2}}}$
Now differentiating with respect to ‘x’.
$\dfrac{d}{{dx}}\left( {\sqrt {\dfrac{1}{{{x^3}}}} } \right) = \dfrac{d}{{dx}}\left( {{x^{ - \dfrac{3}{2}}}} \right)$
$= \dfrac{d}{{dx}}\left( {{x^{ - \dfrac{3}{2}}}} \right)$
We know that $\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}$ . Applying this we have
$= - \dfrac{3}{2}{x^{\left( { - \dfrac{3}{2} - 1} \right)}}$
$= - \dfrac{3}{2}{x^{\left( {\dfrac{{ - 3 - 2}}{2}} \right)}}$
$= - \dfrac{3}{2}{x^{\left( { - \dfrac{5}{2}} \right)}}$
We can stop here but we can still write this in the simplified from. That is sending the ‘x’ term to the denominator we have,
$= - \dfrac{3}{{2{x^{\left( {\dfrac{5}{2}} \right)}}}}$
$= - \dfrac{3}{{2\sqrt {{x^5}} }}$
Thus the derivative of $\sqrt {\dfrac{1}{{{x^3}}}}$ is $- \dfrac{3}{{2\sqrt {{x^5}} }}$ .
So, the correct answer is “$- \dfrac{3}{{2\sqrt {{x^5}} }}$”.

Note : We know that the differentiation of constant is zero. Since the given problem is simple we directly differentiated. But we have different types of differentiation rules.
Sum or difference rule: when the function is the sum or difference of two functions, the derivative is the sum or difference of derivative of each function. i.e.,
$f(x) = u(x) \pm v(x)$ then $f'(x) = u'(x) \pm v'(x)$ .
Product rule: when f(x) is the product of two function that is $f(x) = u(x).v(x)$ then $f'(x) = u'(x).v(x) + u(x).v'(x)$ . We use this rule depending on the given problem.