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Question: How do you differentiate \( 1 + {\cos ^2}(x)? \)...

How do you differentiate 1+cos⁡2(x)?1 + {\cos ^2}(x)?

Explanation

Solution

Hint : Here, we find the first order of the derivative and by using the chain rule of the composite function. First of all we will apply the derivative of the given power in the function and then by using the chain rule will take the derivative of the function and then the derivative of the angle.

Complete step-by-step answer :
Take the given expression –
f(x)=1+cos⁡2(x)f(x) = 1 + {\cos ^2}(x)
The above equation can be re-written as –
f(x)=1+(cos⁡x)2f(x) = 1 + {(\cos x)^2}
Take derivative of the given function –
f′(x)=ddx[1+(cos⁡x)2]f'(x) = \dfrac{d}{{dx}}[1 + {(\cos x)^2}]
Apply derivative separately to both the terms on the right-hand side of the equation.
f′(x)=ddx(1)+ddx(cos⁡x)2f'(x) = \dfrac{d}{{dx}}(1) + \dfrac{d}{{dx}}{(\cos x)^2}
Apply the identity here first of all for xn{x^n} formula in the above equation and then apply (uv)\left( {\dfrac{u}{v}} \right) rule. Also, the derivative of a constant term is always zero.
f′(x)=0+2(cos⁡x)d(cos⁡x)dxf'(x) = 0 + 2(\cos x)\dfrac{{d(\cos x)}}{{dx}}
Simplify the above equation- and place the derivative of cosine as the negative sign.
f′(x)=2(cos⁡x)(−sin⁡x)f'(x) = 2(\cos x)( - \sin x)
Take negative signs outside the bracket.
f′(x)=−2(cos⁡x)(sin⁡x)f'(x) = - 2(\cos x)(\sin x)
Simplify the above equation using the identity, sin⁡2x=2cos⁡xsin⁡x\sin 2x = 2\cos x\sin x
f′(x)=−sin⁡2xf'(x) = - \sin 2x
This is the required solution.
So, the correct answer is “ f′(x)=−sin⁡2xf'(x) = - \sin 2x ”.

Note : Know the difference between the differentiation and the integration and apply formula accordingly. Differentiation can be represented as the rate of change of the function, whereas integration represents the sum of the function over the range. They are inverses of each other.