Answer
385.5k+ views
Hint: We know Chain Rule: $\left( {f(g(x))} \right) = f'(g(x))g'(x)$
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
$y = \cos \left( {\cos x} \right).............................\left( i \right)$
So according to our question we need to find \[\dfrac{{dy}}{{dx}} = \dfrac{{d\cos \left( {\cos x}
\right)}}{{dx}}\]
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:
\[f\left( x \right) = \cos x\,\,{\text{and}}\,\,g\left( x \right) = \cos x.......................\left( {iii} \right)\]
Now to apply chain rule, let:
$u = \cos x$
Then we can say that $f\left( {g\left( x \right)} \right) = \cos u.......................\left( {iv} \right)$
Now on comparing to (ii) we can write:
\[
\dfrac{{dy}}{{dx}} = \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = f'(g(x))g'(x) \\
\Rightarrow \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = \dfrac{d}{{du}}\left( {\cos u}
\right)\dfrac{d}{{dx}}\left( {\cos x} \right)..............................\left( v \right) \\
\]
On observing (v) we can say that the derivative of $\cos u$with respect to $du$would be$ - \sin u$.
So on substituting we get:
\[
\Rightarrow \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = \dfrac{d}{{du}}\left( {\cos u}
\right)\dfrac{d}{{dx}}\left( {\cos x} \right) \\
\Rightarrow \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = - \sin \left( u \right)\dfrac{d}{{dx}}\left(
{\cos x} \right).................................\left( {vi} \right) \\
\]
Substituting $u = \cos x$back to (vi) we get:
\[ \Rightarrow \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = - \sin \left( {\cos x}
\right)\dfrac{d}{{dx}}\left( {\cos x} \right).......................\left( {vii} \right)\]
Also we know that the derivative of $\cos x$with respect to $dx$would be$ - \sin x$.
Substituting this in (vii) we get:
\[
\Rightarrow \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = \left( { - \sin \left( {\cos x} \right)}
\right)\left( { - \sin x} \right) \\
\Rightarrow \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = \left( {\sin \left( {\cos x} \right)}
\right)\left( {\sin x} \right) \\
\]
On rearranging the terms we get:
\[\dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = \sin x\sin \left( {\cos x} \right)........................\left( {viii}
\right)\]
Therefore we can say that the derivative of$y = \cos \left( {\cos x} \right)$ is \[\sin x\sin \left( {\cos x} \right)\].
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
$y = \cos \left( {\cos x} \right).............................\left( i \right)$
So according to our question we need to find \[\dfrac{{dy}}{{dx}} = \dfrac{{d\cos \left( {\cos x}
\right)}}{{dx}}\]
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:
\[f\left( x \right) = \cos x\,\,{\text{and}}\,\,g\left( x \right) = \cos x.......................\left( {iii} \right)\]
Now to apply chain rule, let:
$u = \cos x$
Then we can say that $f\left( {g\left( x \right)} \right) = \cos u.......................\left( {iv} \right)$
Now on comparing to (ii) we can write:
\[
\dfrac{{dy}}{{dx}} = \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = f'(g(x))g'(x) \\
\Rightarrow \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = \dfrac{d}{{du}}\left( {\cos u}
\right)\dfrac{d}{{dx}}\left( {\cos x} \right)..............................\left( v \right) \\
\]
On observing (v) we can say that the derivative of $\cos u$with respect to $du$would be$ - \sin u$.
So on substituting we get:
\[
\Rightarrow \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = \dfrac{d}{{du}}\left( {\cos u}
\right)\dfrac{d}{{dx}}\left( {\cos x} \right) \\
\Rightarrow \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = - \sin \left( u \right)\dfrac{d}{{dx}}\left(
{\cos x} \right).................................\left( {vi} \right) \\
\]
Substituting $u = \cos x$back to (vi) we get:
\[ \Rightarrow \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = - \sin \left( {\cos x}
\right)\dfrac{d}{{dx}}\left( {\cos x} \right).......................\left( {vii} \right)\]
Also we know that the derivative of $\cos x$with respect to $dx$would be$ - \sin x$.
Substituting this in (vii) we get:
\[
\Rightarrow \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = \left( { - \sin \left( {\cos x} \right)}
\right)\left( { - \sin x} \right) \\
\Rightarrow \dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = \left( {\sin \left( {\cos x} \right)}
\right)\left( {\sin x} \right) \\
\]
On rearranging the terms we get:
\[\dfrac{{d\cos \left( {\cos x} \right)}}{{dx}} = \sin x\sin \left( {\cos x} \right)........................\left( {viii}
\right)\]
Therefore we can say that the derivative of$y = \cos \left( {\cos x} \right)$ is \[\sin x\sin \left( {\cos x} \right)\].
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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)