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Calculating Derivative – Deriving an implicit function – Exercise 2119

Exercise

Given the equation:

\cos^2 x + \cos^2 y=\cos (2x+2y)

For

y=f(x)

Find the derivative of the following function:

y'=f'(x)

Final Answer


f'(x)=\frac{\cos x\sin x-\sin(2x+2y)}{\sin(2x+2y)-\cos y\sin y}

Solution

We need to calculate a derivative of an implicit function. To do that, we use the chain rule in Derivative Rules on both sides of the equation and when we encounter y, we derive it, i.e. multiply by y’. This is how it goes:

\cos^2 x + \cos^2 y=\cos (2x+2y)

2\cos x\cdot (-\sin x)+2\cos y\cdot(-\sin y)\cdot y'=-\sin(2x+2y)\cdot(2+2y')

Now, we want to isolate y’. Therefore, we open brackets and move expressions with y’ to one side:

-2\cos x\sin x-2y'\cos y\sin y=-2\sin(2x+2y)-2y'\sin(2x+2y)

2y'\sin (2x+2y)-2y'\cos y\sin y=-2\sin (2x+2y)+2\cos x\sin x

y'(2\sin(2x+2y)-2\cos y\sin y)=2\cos x\sin x-2\sin (2x+2y)

y'=\frac{2\cos x\sin x-2\sin(2x+2y)}{2\sin(2x+2y)-2\cos y\sin y}

y'=\frac{\cos x\sin x-\sin(2x+2y)}{\sin(2x+2y)-\cos y\sin y}

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