Answer
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Hint: We are given a trigonometric expression which has inverse function in it and has the value of ‘x’ in the given range. And in the question, we are asked to find the derivative of the given expression. We will first let the given expression be equal to a variable say ‘y’. We will begin by simplifying the expression as much as possible and the new expression that we will get is, \[y=\dfrac{\pi }{4}-x\]. Now, we will differentiate the obtained expression with respect to ‘x’. Hence, we will have the required derivative of the given expression.
Complete step-by-step solution:
According to the given question, we are given an expression based on trigonometric functions and we are asked to find the derivative of the same.
The expression that we have is,
\[{{\cos }^{-1}}\left( \dfrac{\sin x+\cos x}{\sqrt{2}} \right)\]
Firstly, we let the given expression be equal to ‘y’, that is,
\[y={{\cos }^{-1}}\left( \dfrac{\sin x+\cos x}{\sqrt{2}} \right)\]
Now, we will simplify the above given expression first as much as possible using the trigonometric identities wherever applicable.
We will now the separate the terms within the bracket, we get,
\[\Rightarrow y={{\cos }^{-1}}\left( \dfrac{\sin x}{\sqrt{2}}+\dfrac{\cos x}{\sqrt{2}} \right)\]
We can write the above equation as following,
\[\Rightarrow y={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}}\sin x+\dfrac{1}{\sqrt{2}}\cos x \right)\]
we know the trigonometric identity which goes like, \[\cos (A-B)=\cos A\cos B+\sin A\sin B\], so we will write \[\dfrac{1}{\sqrt{2}}\] using the sine and cosine function such that the expression is similar to the identity.
So, we have,
\[\Rightarrow y={{\cos }^{-1}}\left( \sin \dfrac{\pi }{4}\sin x+\cos \dfrac{\pi }{4}\cos x \right)\]
Rearranging the above expression, we get,
\[\Rightarrow y={{\cos }^{-1}}\left( \cos \dfrac{\pi }{4}\cos x+\sin \dfrac{\pi }{4}\sin x \right)\]
Now, we have the expression similar to, \[\cos (A-B)=\cos A\cos B+\sin A\sin B\], so we can write the expression as,
\[\Rightarrow y={{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{4}-x \right) \right)\]
We know that, \[{{\cos }^{-1}}\left( \cos \theta \right)=\theta \], we get,
\[\Rightarrow y=\dfrac{\pi }{4}-x\]
We now have the simplified form of the given expression so now we will differentiate the given expression and we get,
\[\dfrac{dy}{dx}=-1\]
Therefore, the derivative of the given expression is -1.
Note: The expression given to us should not be straightaway to find the derivative as then the differentiation will prove to be very complicated and lengthy. So, our first step should always be to simplify the given expression, if possible and only then carry out the differentiation of the expression. Also, the trigonometric identities should be known beforehand, else the expression won’t get solved easily.
Complete step-by-step solution:
According to the given question, we are given an expression based on trigonometric functions and we are asked to find the derivative of the same.
The expression that we have is,
\[{{\cos }^{-1}}\left( \dfrac{\sin x+\cos x}{\sqrt{2}} \right)\]
Firstly, we let the given expression be equal to ‘y’, that is,
\[y={{\cos }^{-1}}\left( \dfrac{\sin x+\cos x}{\sqrt{2}} \right)\]
Now, we will simplify the above given expression first as much as possible using the trigonometric identities wherever applicable.
We will now the separate the terms within the bracket, we get,
\[\Rightarrow y={{\cos }^{-1}}\left( \dfrac{\sin x}{\sqrt{2}}+\dfrac{\cos x}{\sqrt{2}} \right)\]
We can write the above equation as following,
\[\Rightarrow y={{\cos }^{-1}}\left( \dfrac{1}{\sqrt{2}}\sin x+\dfrac{1}{\sqrt{2}}\cos x \right)\]
we know the trigonometric identity which goes like, \[\cos (A-B)=\cos A\cos B+\sin A\sin B\], so we will write \[\dfrac{1}{\sqrt{2}}\] using the sine and cosine function such that the expression is similar to the identity.
So, we have,
\[\Rightarrow y={{\cos }^{-1}}\left( \sin \dfrac{\pi }{4}\sin x+\cos \dfrac{\pi }{4}\cos x \right)\]
Rearranging the above expression, we get,
\[\Rightarrow y={{\cos }^{-1}}\left( \cos \dfrac{\pi }{4}\cos x+\sin \dfrac{\pi }{4}\sin x \right)\]
Now, we have the expression similar to, \[\cos (A-B)=\cos A\cos B+\sin A\sin B\], so we can write the expression as,
\[\Rightarrow y={{\cos }^{-1}}\left( \cos \left( \dfrac{\pi }{4}-x \right) \right)\]
We know that, \[{{\cos }^{-1}}\left( \cos \theta \right)=\theta \], we get,
\[\Rightarrow y=\dfrac{\pi }{4}-x\]
We now have the simplified form of the given expression so now we will differentiate the given expression and we get,
\[\dfrac{dy}{dx}=-1\]
Therefore, the derivative of the given expression is -1.
Note: The expression given to us should not be straightaway to find the derivative as then the differentiation will prove to be very complicated and lengthy. So, our first step should always be to simplify the given expression, if possible and only then carry out the differentiation of the expression. Also, the trigonometric identities should be known beforehand, else the expression won’t get solved easily.
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