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Question

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

How do you differentiate y=sin2x+cos2xy = {\sin ^2}x + {\cos ^2}x?

Explanation

Solution

According to the question we have to determine the differentiation of the trigonometric expression which is y=sin2x+cos2xy = {\sin ^2}x + {\cos ^2}x as mentioned in the question. So, first of all to determine the required differentiation of the trigonometric expression we have to use the formula to find the differentiation which is as mentioned below:

Formula used:
dxndx=nxn1...............(A)\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}...............(A)
Now, to find the differentiation of the trimetric term which is sinx\sin xwe have to use the formula as mentioned below:

dsinxdx=cosx...............(B) \dfrac{{d\sin x}}{{dx}} = \cos x...............(B)
Now, we have to determine the differentiation of an-other remaining trigonometric term which is cosx\cos xcan be determined with the help of the formula which is as mentioned below:

dcosxdx=sinx...............(C) \dfrac{{d\cos x}}{{dx}} = - \sin x...............(C)
Now, to solve the expression obtained we just have to add or subtract the terms which can be added or subtracted.

Complete step by step solution:
Step 1: First of all to determine the required differentiation of the trigonometric expression we have to use the formula (A) to find the differentiation which is as mentioned in the solution hint. Hence,
dydx=2(sinx)21dsinxdx+2(cosx)21dcosxdx\Rightarrow \dfrac{{dy}}{{dx}} = 2{(\sin x)^{2 - 1}}\dfrac{{d\sin x}}{{dx}} + 2{(\cos x)^{2 - 1}}\dfrac{{d\cos x}}{{dx}}
Step 2: Now, to find the differentiation of the trimetric term which is sinx\sin xwe have to use the formula (B) as mentioned in the solution hint. Hence,
dydx=2sinx(cosx)+2(cosx)dcosxdx\Rightarrow \dfrac{{dy}}{{dx}} = 2\sin x(\cos x) + 2(\cos x)\dfrac{{d\cos x}}{{dx}}
Step 3: Now, we have to determine the differentiation of an-other remaining trigonometric term which is cosx\cos xcan be determined with the help of the formula (C) which is as mentioned in the solution hint. Hence,
dydx=2sinx(cosx)+2cosx(sinx)\Rightarrow \dfrac{{dy}}{{dx}} = 2\sin x(\cos x) + 2\cos x( - \sin x)
Step 4: Now, to solve the expression obtained we just have to add or subtract the terms which can be added or subtracted. Hence,
dydx=2sinxcosx2cosxsinx dydx=0  \Rightarrow \dfrac{{dy}}{{dx}} = 2\sin x\cos x - 2\cos x\sin x \\\ \Rightarrow \dfrac{{dy}}{{dx}} = 0 \\\
Final solution: Hence, with the help of the formulas (A), (B) and (C) we have determined the required differentiation which is 0.

Note:

  1. To determine the differentiation of the term of the given trigonometric function which are sin2x{\sin ^2}x and cos2x{\cos ^2}x we have to usedxndx=nxn1\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}the trigonometric function easily.
  2. To obtain the differentiation of the trigonometric function which are sinx\sin xand cosx\cos xwe have to use the formula to find the differentiation of these trigonometric terms which are mentioned in the solution hint.