Differentiate ${\sin ^3}x.$
Answer
624.3k+ 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
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

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

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

