
Differentiate ${\sin ^3}x.$
Answer
557.1k+ views
Hint:We know Chain Rule: \[f(g(x)) = f'(g(x))g'(x)\]
By using Chain rule we can solve this problem.
Since we cannot find the direct derivative of the given question we have to use the chain rule. So we must convert our question in the form of the equation above such that we have to find the values of every term in the above equation and substitute it back. In that way we would be able to find the solution for the given question.
Complete step by step solution:
Given
${\sin ^3}x.............................\left( i \right)$
So according to our question we need to find \[\dfrac{{d{{\sin }^3}x}}{{dx}}.\]
Thus here we can use chain rule to find the derivative since we can’t find the derivative with any direct equation.
Now we know that chain rule is:\[f(g(x)) = f'(g(x))g'(x).......................\left( {ii} \right)\]
Such that on comparing (ii), if:
$f\left( x \right) = {x^3}\,\,{\text{and}}\,\,g\left( x \right) = \sin x.......................\left( {iii} \right)$
Then we can say that $f\left( {g\left( x \right)} \right) = {\left( {\sin x} \right)^3}.......................\left(
{iv} \right)$
Now we have to find:
$f'\left( {g\left( x \right)} \right)\,\,and\,\,g'\left( x \right)$
So using (iii) and (iv) to find$f'\left( {g\left( x \right)} \right)\,\,and\,\,g'\left( x \right)$:
By using (iv) we can write:
$
f\left( {g\left( x \right)} \right) = {\left( {\sin x} \right)^3} \\
\Rightarrow f'\left( {g\left( x \right)} \right) = 3{\sin ^2}x...................\left( v \right) \\
$
And by using (iii) we can write:
$
g\left( x \right) = \sin x \\
\Rightarrow g'\left( x \right) = \cos x.........................\left( {vi} \right) \\
$
Now substituting (v) and (vi) in (ii), we get:
\[
f(g(x)) = f'(g(x))g'(x) \\
\Rightarrow f(g(x)) = 3{\sin ^2}x\cos x..............\left( {vii} \right) \\
\]
Now we know that \[f(g(x))\]is our required derivative that we need to find such that:
\[f(g(x)) = \dfrac{{d{{\sin }^3}x}}{{dx}}\]
Therefore we can write our final answer as:
\[\dfrac{{d{{\sin }^3}x}}{{dx}} = 3{\sin ^2}x\cos x\]
Note:
The Chain Rule can also be written as:
$\dfrac{{df}}{{dx}} = \dfrac{{df}}{{dg}} \times \dfrac{{dg}}{{dh}} \times \dfrac{{dh}}{{dx}}$
It mainly tells us how to differentiate composite functions. 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.
By using Chain rule we can solve this problem.
Since we cannot find the direct derivative of the given question we have to use the chain rule. So we must convert our question in the form of the equation above such that we have to find the values of every term in the above equation and substitute it back. In that way we would be able to find the solution for the given question.
Complete step by step solution:
Given
${\sin ^3}x.............................\left( i \right)$
So according to our question we need to find \[\dfrac{{d{{\sin }^3}x}}{{dx}}.\]
Thus here we can use chain rule to find the derivative since we can’t find the derivative with any direct equation.
Now we know that chain rule is:\[f(g(x)) = f'(g(x))g'(x).......................\left( {ii} \right)\]
Such that on comparing (ii), if:
$f\left( x \right) = {x^3}\,\,{\text{and}}\,\,g\left( x \right) = \sin x.......................\left( {iii} \right)$
Then we can say that $f\left( {g\left( x \right)} \right) = {\left( {\sin x} \right)^3}.......................\left(
{iv} \right)$
Now we have to find:
$f'\left( {g\left( x \right)} \right)\,\,and\,\,g'\left( x \right)$
So using (iii) and (iv) to find$f'\left( {g\left( x \right)} \right)\,\,and\,\,g'\left( x \right)$:
By using (iv) we can write:
$
f\left( {g\left( x \right)} \right) = {\left( {\sin x} \right)^3} \\
\Rightarrow f'\left( {g\left( x \right)} \right) = 3{\sin ^2}x...................\left( v \right) \\
$
And by using (iii) we can write:
$
g\left( x \right) = \sin x \\
\Rightarrow g'\left( x \right) = \cos x.........................\left( {vi} \right) \\
$
Now substituting (v) and (vi) in (ii), we get:
\[
f(g(x)) = f'(g(x))g'(x) \\
\Rightarrow f(g(x)) = 3{\sin ^2}x\cos x..............\left( {vii} \right) \\
\]
Now we know that \[f(g(x))\]is our required derivative that we need to find such that:
\[f(g(x)) = \dfrac{{d{{\sin }^3}x}}{{dx}}\]
Therefore we can write our final answer as:
\[\dfrac{{d{{\sin }^3}x}}{{dx}} = 3{\sin ^2}x\cos x\]
Note:
The Chain Rule can also be written as:
$\dfrac{{df}}{{dx}} = \dfrac{{df}}{{dg}} \times \dfrac{{dg}}{{dh}} \times \dfrac{{dh}}{{dx}}$
It mainly tells us how to differentiate composite functions. 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
Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

10 examples of friction in our daily life

