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.