How to differentiate \[\log \left( \cos ec\left( x \right)-\cot \left( x \right) \right)\]?
Answer
615.3k+ views
Hint: In the given question, we have been asked to differentiate the function. The given equation is based on the concepts of derivatives and trigonometry. In order to differentiate the given function, we will apply the identities of trigonometric and the identities of derivation as well. First we need to apply the chain rule of derivation and later on in the solution, we would need to apply the properties of differentiation. And we will get the required solution.
Complete step by step solution:
We have given,
\[\Rightarrow \log \left( \cos ec\left( x \right)-\cot \left( x \right) \right)\]
Suppose \[y=\log \left( \cos ec\left( x \right)-\cot \left( x \right) \right)\]
Differentiating the y with respect to x, we get
\[\dfrac{dy}{dx}=\dfrac{d}{dx}\log \left( \cos ec\left( x \right)-\cot \left( x \right) \right)\]
On using chain rule, we obtain
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos ec\left( x \right)-\cot \left( x \right)}\left( \dfrac{d}{dx}\left( \cos ec\left( x \right)-\cot \left( x \right) \right) \right)\]
Simplifying the above, we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos ec\left( x \right)-\cot \left( x \right)}\left( \dfrac{d}{dx}\cos ec\left( x \right)-\dfrac{d}{dx}\cot \left( x \right) \right)\]
Using the differentiation of trigonometric identities,
\[\dfrac{d}{dx}\log x=\dfrac{1}{x}\]
\[\dfrac{d}{dx}\cos ecx=-\cos ec\left( x \right)\times \cot \left( x \right)\]
\[\dfrac{d}{dx}\cot x=-\cos e{{c}^{2}}\left( x \right)\]
Applying the identities, we obtain
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos ec\left( x \right)-\cot \left( x \right)}\left( -\cos ec\left( x \right)\times \cot \left( x \right)-\left( -\cos e{{c}^{2}}\left( x \right) \right) \right)\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos ec\left( x \right)-\cot \left( x \right)}\left( -\cos ec\left( x \right)\times \cot \left( x \right)+\cos e{{c}^{2}}\left( x \right) \right)\]
Rearranging the terms, we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos ec\left( x \right)-\cot \left( x \right)}\left( \cos e{{c}^{2}}x-\cos ec\left( x \right)\cot \left( x \right) \right)\]
Taking cosec(x) common from the bracket, we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos ec\left( x \right)-\cot \left( x \right)}\left( \cos ec\left( x \right)\left( \cos ecx-\cot x \right) \right)\]
Simplifying the above, we get
\[\Rightarrow \dfrac{dy}{dx}=\cos ec\left( x \right)\]
\[\therefore \dfrac{d}{dx}\log \left( \cos ecx-\cot x \right)=\cos ec\left( x \right)\]
Therefore, this is the required differentiation.
Note: In these types of question of differentiation, first we need to apply the chain rule of differentiation. There are other method also for differentiating a given function such as product rule of differentiation etc. we should always remember the properties of derivation of trigonometric function, that would make question easier to solve. We should perform each step carefully in order to avoid confusion otherwise it will result in making errors. So always examine explicitly to avoid calculation method while solving the question. We should always know the basic standard deviation formulas as we can use these formulas directly.
Complete step by step solution:
We have given,
\[\Rightarrow \log \left( \cos ec\left( x \right)-\cot \left( x \right) \right)\]
Suppose \[y=\log \left( \cos ec\left( x \right)-\cot \left( x \right) \right)\]
Differentiating the y with respect to x, we get
\[\dfrac{dy}{dx}=\dfrac{d}{dx}\log \left( \cos ec\left( x \right)-\cot \left( x \right) \right)\]
On using chain rule, we obtain
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos ec\left( x \right)-\cot \left( x \right)}\left( \dfrac{d}{dx}\left( \cos ec\left( x \right)-\cot \left( x \right) \right) \right)\]
Simplifying the above, we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos ec\left( x \right)-\cot \left( x \right)}\left( \dfrac{d}{dx}\cos ec\left( x \right)-\dfrac{d}{dx}\cot \left( x \right) \right)\]
Using the differentiation of trigonometric identities,
\[\dfrac{d}{dx}\log x=\dfrac{1}{x}\]
\[\dfrac{d}{dx}\cos ecx=-\cos ec\left( x \right)\times \cot \left( x \right)\]
\[\dfrac{d}{dx}\cot x=-\cos e{{c}^{2}}\left( x \right)\]
Applying the identities, we obtain
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos ec\left( x \right)-\cot \left( x \right)}\left( -\cos ec\left( x \right)\times \cot \left( x \right)-\left( -\cos e{{c}^{2}}\left( x \right) \right) \right)\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos ec\left( x \right)-\cot \left( x \right)}\left( -\cos ec\left( x \right)\times \cot \left( x \right)+\cos e{{c}^{2}}\left( x \right) \right)\]
Rearranging the terms, we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos ec\left( x \right)-\cot \left( x \right)}\left( \cos e{{c}^{2}}x-\cos ec\left( x \right)\cot \left( x \right) \right)\]
Taking cosec(x) common from the bracket, we get
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\cos ec\left( x \right)-\cot \left( x \right)}\left( \cos ec\left( x \right)\left( \cos ecx-\cot x \right) \right)\]
Simplifying the above, we get
\[\Rightarrow \dfrac{dy}{dx}=\cos ec\left( x \right)\]
\[\therefore \dfrac{d}{dx}\log \left( \cos ecx-\cot x \right)=\cos ec\left( x \right)\]
Therefore, this is the required differentiation.
Note: In these types of question of differentiation, first we need to apply the chain rule of differentiation. There are other method also for differentiating a given function such as product rule of differentiation etc. we should always remember the properties of derivation of trigonometric function, that would make question easier to solve. We should perform each step carefully in order to avoid confusion otherwise it will result in making errors. So always examine explicitly to avoid calculation method while solving the question. We should always know the basic standard deviation formulas as we can use these formulas directly.
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