
How do you use the chain rule to differentiate \[y={{\sin }^{2}}\left( \cos \left( 4x \right) \right)\]?
Answer
547.8k+ views
Hint: To differentiate, we have to use chain rule which states that the derivative of \[f\left( g\left( x \right) \right)\] is \[f'\left( g\left( x \right) \right).g'\left( x \right)\]. Also we use chain rules to differentiate composite functions. Composite function is created when \[1\] function is substituted in another example of composite function \[f\left( g\left( x \right) \right)\].
Complete Step by Step solution:
Here, we have \[y={{\sin }^{2}}\left( \cos \left( 4x \right) \right)\]
Let \[f\left( x \right)={{\sin }^{2}}\left( x \right)\] and \[g\left( x \right)=\cos \left( 4x \right)\]
Thus, we get \[y=f\left( g\left( x \right) \right)={{\sin }^{2}}\left( \cos \left( 4x \right) \right)\]
Now, we can observe that the above function is a composite function.
So, here we can apply the chain rule.
\[\therefore \] By chain rule we get,
\[\dfrac{df\left( g\left( x \right) \right)}{dx}=f'\left( g\left( x \right) \right).g'\left( x \right)\]
\[=2\sin \left( \cos \left( 4x \right) \right).cos\left( \cos \left( 4x \right) \right)\]
\[\Rightarrow 2\sin \left( \cos \left( 4x \right) \right).cos\left( \cos \left( 4x \right) \right)\] has a composite function as \[\cos \left( \cos \left( 4x \right) \right)\]
Thus, by applying chain rule again we get,
\[\dfrac{dy}{dx}=2\sin \left( \cos \left( 4x \right) \right).cos\left( \cos \left( 4x \right) \right).\dfrac{d}{dx}\cos \left( 4x \right)\]
Here, again we can observe that, \[\cos \left( 4x \right)\] is a composite function. So, we will again apply the chain rule, we get,
\[\dfrac{dy}{dx}=2\sin \left( \cos \left( 4x \right) \right).cos\left( \cos \left( 4x \right) \right).\left( -\sin \left( 4x \right) \right).4\]
\[=-8\sin \left( 4x \right)\sin \left( \cos \left( 4x \right) \right)\cos \left( \cos \left( 4x \right) \right)\]
Note:
In differentiating such problems after solving the composite function we can again get the composite function thus, we have to keep simplifying the composite functions by chain rule. In case if the composite function continues then here we use the concept of ‘partial differentiation’. By this it can be simplified.
Complete Step by Step solution:
Here, we have \[y={{\sin }^{2}}\left( \cos \left( 4x \right) \right)\]
Let \[f\left( x \right)={{\sin }^{2}}\left( x \right)\] and \[g\left( x \right)=\cos \left( 4x \right)\]
Thus, we get \[y=f\left( g\left( x \right) \right)={{\sin }^{2}}\left( \cos \left( 4x \right) \right)\]
Now, we can observe that the above function is a composite function.
So, here we can apply the chain rule.
\[\therefore \] By chain rule we get,
\[\dfrac{df\left( g\left( x \right) \right)}{dx}=f'\left( g\left( x \right) \right).g'\left( x \right)\]
\[=2\sin \left( \cos \left( 4x \right) \right).cos\left( \cos \left( 4x \right) \right)\]
\[\Rightarrow 2\sin \left( \cos \left( 4x \right) \right).cos\left( \cos \left( 4x \right) \right)\] has a composite function as \[\cos \left( \cos \left( 4x \right) \right)\]
Thus, by applying chain rule again we get,
\[\dfrac{dy}{dx}=2\sin \left( \cos \left( 4x \right) \right).cos\left( \cos \left( 4x \right) \right).\dfrac{d}{dx}\cos \left( 4x \right)\]
Here, again we can observe that, \[\cos \left( 4x \right)\] is a composite function. So, we will again apply the chain rule, we get,
\[\dfrac{dy}{dx}=2\sin \left( \cos \left( 4x \right) \right).cos\left( \cos \left( 4x \right) \right).\left( -\sin \left( 4x \right) \right).4\]
\[=-8\sin \left( 4x \right)\sin \left( \cos \left( 4x \right) \right)\cos \left( \cos \left( 4x \right) \right)\]
Note:
In differentiating such problems after solving the composite function we can again get the composite function thus, we have to keep simplifying the composite functions by chain rule. In case if the composite function continues then here we use the concept of ‘partial differentiation’. By this it can be simplified.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

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

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

