# What is the derivative of # f(x)=secx^2cos^2x#?

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To find the derivative of (f(x) = \sec^2(x) \cos^2(x)), you can use the product rule and the chain rule. Here's the solution:

Given: [ f(x) = \sec^2(x) \cos^2(x) ]

Using the product rule: [ \frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x) ]

Let: [ u(x) = \sec^2(x) ] [ v(x) = \cos^2(x) ]

Find derivatives: [ u'(x) = 2\sec(x)\tan(x)\sec(x) ] [ v'(x) = -2\cos(x)\sin(x) ]

Apply the product rule: [ f'(x) = u'(x)v(x) + u(x)v'(x) ] [ f'(x) = (2\sec(x)\tan(x)\sec(x))(\cos^2(x)) + (\sec^2(x))(-2\cos(x)\sin(x)) ]

[ f'(x) = 2\sec(x)\tan(x)\sec(x)\cos^2(x) - 2\sec^2(x)\cos(x)\sin(x) ]

[ f'(x) = 2\sec^3(x)\tan(x)\cos(x) - 2\sec^2(x)\cos(x)\sin(x) ]

So, the derivative of ( f(x) = \sec^2(x) \cos^2(x) ) is: [ f'(x) = 2\sec^3(x)\tan(x)\cos(x) - 2\sec^2(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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