Answer
Verified
410.4k+ views
Hint: 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:
\[ - \left( {\dfrac{1}{2}} \right){\left( {x - 1} \right)^{ - \left( {\dfrac{3}{2}} \right)}} = - \left(
{\dfrac{1}{{2\sqrt {{{\left( {x - 1} \right)}^3}} }}} \right)....................\left( {viii} \right)\]
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.
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:
\[ - \left( {\dfrac{1}{2}} \right){\left( {x - 1} \right)^{ - \left( {\dfrac{3}{2}} \right)}} = - \left(
{\dfrac{1}{{2\sqrt {{{\left( {x - 1} \right)}^3}} }}} \right)....................\left( {viii} \right)\]
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE