# How do you find the derivative of #sin^2x+cos^2x#?

Using the Pythagorean identity is the simplest method:

Let's say we are aware of:

Rule of power:

Chain principle:

Then:

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To find the derivative of ( \sin^2(x) + \cos^2(x) ), you can use the chain rule and the trigonometric identities.

Here's the calculation:

[ \frac{d}{dx}(\sin^2(x) + \cos^2(x)) = \frac{d}{dx}(\sin^2(x)) + \frac{d}{dx}(\cos^2(x)). ]

Using the chain rule, we get:

[ \frac{d}{dx}(\sin^2(x)) = 2\sin(x)\cos(x) \frac{d}{dx}(\sin(x)) ] [ \frac{d}{dx}(\cos^2(x)) = 2\sin(x)\cos(x) \frac{d}{dx}(\cos(x)). ]

Now, ( \frac{d}{dx}(\sin(x)) = \cos(x) ) and ( \frac{d}{dx}(\cos(x)) = -\sin(x) ).

Substituting these derivatives, we get:

[ \frac{d}{dx}(\sin^2(x) + \cos^2(x)) = 2\sin(x)\cos(x) (\cos(x)) + 2\sin(x)\cos(x) (-\sin(x)). ]

This simplifies to:

[ 2\sin(x)\cos(x)(\cos(x) - \sin(x)). ]

So, the derivative of ( \sin^2(x) + \cos^2(x) ) with respect to ( x ) is ( 2\sin(x)\cos(x)(\cos(x) - \sin(x)) ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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