Question

Find the derivative of $\dfrac{1}{{\sqrt {x - 1} }}$.

Answer 1

We know that the exponential property: $\sqrt x = {\left( x \right)^{\dfrac{1}{2}}}$so we can convert our given question in the form of the above given identity and there by simplify it. Also to find the derivative we have the formula:$\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$ So by using the above equations and identities we can simplify the given question.

Complete step by step solution:
Given
$\dfrac{1}{{\sqrt {x - 1} }}....................\left( i \right)$
We need to find the derivative of (i), such that:
$\dfrac{d}{{dx}}\left( {\dfrac{1}{{\sqrt {x - 1} }}} \right)...........................\left( {ii} \right)$
Now we know the exponential identity$\sqrt x = {\left( x \right)^{\dfrac{1}{2}}}$, such that on applying it to (i) we get:
$\Rightarrow \sqrt {x - 1} = {\left( {x - 1} \right)^{\dfrac{1}{2}}} \\\ \Rightarrow \dfrac{1}{{\sqrt {x - 1} }} = \dfrac{1}{{{{\left( {x - 1} \right)}^{\dfrac{1}{2}}}}}.....................\left( {iii} \right) \\\$
Now we know another exponential identity:
$\dfrac{1}{{{x^n}}} = {x^{ - n}}.........................\left( {iv} \right)$
Applying (iv) on (iii) we get:
$\dfrac{1}{{{{\left( {x - 1} \right)}^{\dfrac{1}{2}}}}} = {\left( {x - 1} \right)^{ - \left( {\dfrac{1}{2}} \right)}}...................\left( v \right)$

Now we have to substitute (v) in (ii), and thus have to find the derivative.

On substituting we get:
$\dfrac{d}{{dx}}\left( {\dfrac{1}{{\sqrt {x - 1} }}} \right) = \dfrac{d}{{dx}}{\left( {x - 1} \right)^{ - \left( {\dfrac{1}{2}} \right)}}............\left( {vi} \right)$

Now to solve (vi) we have the basic identity to find the derivative of ${x^n}$ as:
$\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$
But here we don’t have ${x^n}$but we have ${\left( {x - 1} \right)^n}$so here we should apply chain rule:
On comparing the above equation we can say that here$n = - \left( {\dfrac{1}{2}} \right)$.
Now applying the identity on (vii) we get:
$\dfrac{d}{{dx}}{\left( {x - 1} \right)^{ - \left( {\dfrac{1}{2}} \right)}} = - \left( {\dfrac{1}{2}} \right){\left( {x - 1} \right)^{\left( { - \dfrac{1}{2} - 1} \right)}} \times \dfrac{d}{{dx}}\left( {x - 1} \right) \\\ = - \left( {\dfrac{1}{2}} \right){\left( {x - 1} \right)^{ - \left( {\dfrac{3}{2}} \right)}} \times 1 \\\ = - \left( {\dfrac{1}{2}} \right){\left( {x - 1} \right)^{ - \left( {\dfrac{3}{2}} \right)}}...........................\left( {vii} \right) \\\$
Now using the same property $\sqrt x = {\left( x \right)^{\dfrac{1}{2}}}$and $\dfrac{1}{{{x^n}}} = {x^{ - n}}$we can write (vii) as below:

{\dfrac{1}{{2\sqrt {{{\left( {x - 1} \right)}^3}} }}} \right)....................\left( {viii} \right)$$ Therefore the derivative of$\dfrac{1}{{\sqrt {x - 1} }}$is$$ - \left( {\dfrac{1}{{2\sqrt {{{\left( {x - 1} \right)}^3}} }}} \right)$$. **Note:** Whenever questions including exponents are given some of the identities useful are: $ {x^m} \times {x^n} = {\left( x \right)^{m + n}} \\\ \dfrac{{{x^n}}}{{{x^m}}} = {\left( x \right)^{n - m}} \\\ {\left( {{x^n}} \right)^m} = {\left( x \right)^{n \times m}} \\\ $ So our given expressions should be converted and expressed based on the above standard identities, by which it would be much easier to simplify and solve it. The Chain Rule can be written as: $\left( {f(g(x))} \right) = f'(g(x))g'(x)$ Chain rule is mainly used for finding the derivative of a composite function. Also care must be taken while using chain rule since it should be applied only on composite functions and applying chain rule that isn’t composite may result in a wrong derivative.