
What is the derivative of \[y=\cos \left( \cos \left( \cos x \right) \right)\]?
Answer
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Hint: We are given a question to find the derivative of the given function. In order to find the derivative of the given function, we will use the chain rule. Firstly, we will differentiate the outermost cosine function. We know that the differentiation of the cosine function is, \[\dfrac{d}{dx}\left( \cos x \right)=-\sin x\]. Then, we will proceed onto differentiating the inner cosine functions. Hence, we will have the derivative if the given function.
Complete step-by-step answer:
According to the given question, we are given a trigonometric function which we have to differentiate and find the value.
We will find the derivative of the given function by using the chain rule.
Chain rule states that if we have to differentiate a function ‘y’, we can the function further and have the main function differentiated as components, that is,
\[\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}\]
The given function that we have is,
\[y=\cos \left( \cos \left( \cos x \right) \right)\]
We will differentiate the above function with respect to ‘x’. Also, we know that the derivative of cosine function is, \[\dfrac{d}{dx}\left( \cos x \right)=-\sin x\].
We get,
\[\Rightarrow \dfrac{dy}{dx}=-\sin x\left( \cos \left( \cos x \right) \right)\times \dfrac{d}{dx}\left( \cos \left( \cos x \right) \right)\]
We started with differentiating the outermost cosine function and then we will proceed onto differentiating the inner cosine functions. Differentiating further, we get,
\[\Rightarrow \dfrac{dy}{dx}=-\sin x\left( \cos \left( \cos x \right) \right)\times \left( -\sin \left( \cos x \right) \right)\times \dfrac{d}{dx}\left( \cos x \right)\]
We will now differentiate the innermost cosine function and we have,
\[\Rightarrow \dfrac{dy}{dx}=-\sin x\left( \cos \left( \cos x \right) \right)\times \left( -\sin \left( \cos x \right) \right)\times \left( -\sin x \right)\]
We now have three negative signs in the above expression. Two of the negative signs will cancel out each other and one will be left. So, we get,
\[\Rightarrow \dfrac{dy}{dx}=-\sin x\left( \cos \left( \cos x \right) \right)\sin \left( \cos x \right)\sin x\]
Therefore, \[\dfrac{dy}{dx}=-\sin x\left( \cos \left( \cos x \right) \right)\sin \left( \cos x \right)\sin x\]
Note: The given function is a composite function, that is, a function within a function. While differentiating a composite function always make sure that the differentiation is carried out from the outermost function towards the innermost one. It should be done the other way round because then the question itself will get complex and you won’t get the answer correct. Also, the derivative of cosine function has negative function along with the sine function. Do not forget the negative sign.
Complete step-by-step answer:
According to the given question, we are given a trigonometric function which we have to differentiate and find the value.
We will find the derivative of the given function by using the chain rule.
Chain rule states that if we have to differentiate a function ‘y’, we can the function further and have the main function differentiated as components, that is,
\[\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dx}\]
The given function that we have is,
\[y=\cos \left( \cos \left( \cos x \right) \right)\]
We will differentiate the above function with respect to ‘x’. Also, we know that the derivative of cosine function is, \[\dfrac{d}{dx}\left( \cos x \right)=-\sin x\].
We get,
\[\Rightarrow \dfrac{dy}{dx}=-\sin x\left( \cos \left( \cos x \right) \right)\times \dfrac{d}{dx}\left( \cos \left( \cos x \right) \right)\]
We started with differentiating the outermost cosine function and then we will proceed onto differentiating the inner cosine functions. Differentiating further, we get,
\[\Rightarrow \dfrac{dy}{dx}=-\sin x\left( \cos \left( \cos x \right) \right)\times \left( -\sin \left( \cos x \right) \right)\times \dfrac{d}{dx}\left( \cos x \right)\]
We will now differentiate the innermost cosine function and we have,
\[\Rightarrow \dfrac{dy}{dx}=-\sin x\left( \cos \left( \cos x \right) \right)\times \left( -\sin \left( \cos x \right) \right)\times \left( -\sin x \right)\]
We now have three negative signs in the above expression. Two of the negative signs will cancel out each other and one will be left. So, we get,
\[\Rightarrow \dfrac{dy}{dx}=-\sin x\left( \cos \left( \cos x \right) \right)\sin \left( \cos x \right)\sin x\]
Therefore, \[\dfrac{dy}{dx}=-\sin x\left( \cos \left( \cos x \right) \right)\sin \left( \cos x \right)\sin x\]
Note: The given function is a composite function, that is, a function within a function. While differentiating a composite function always make sure that the differentiation is carried out from the outermost function towards the innermost one. It should be done the other way round because then the question itself will get complex and you won’t get the answer correct. Also, the derivative of cosine function has negative function along with the sine function. Do not forget the negative sign.
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