How do you find the derivative of #[3 cos 2x + sin^2 x]#?

Answer 1

The derivative is #-4 sin(x) cos(x) #.

The chain rule must be applied when differentiating:

if #f(x) = (u @ v)(x) = u(v(x))#, then #f'(x) = u'(v(x)) * v'(x)#.

By applying this rule, you can calculate the derivative in the following way:

#f'(x) = 3 * (-sin(2x)) * 2 + 2 * sin(x) * cos(x)#
#color(white)(xxxx) = -6 sin(2x) + 2 sin(x) cos(x)#
... use the transformation #sin(2x) = 2 sin(x) cos(x)# ...
#color(white)(xxxx) = -12 sin(x) cos(x) + 2 sin(x) cos(x)#
#color(white)(xxxx) = -10 sin(x) cos(x) #
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Answer 2

To find the derivative of [3 cos 2x + sin^2 x], you differentiate each term individually using the appropriate rules of differentiation:

  1. Differentiate 3 cos 2x: d/dx [3 cos 2x] = -6 sin 2x

  2. Differentiate sin^2 x: d/dx [sin^2 x] = 2 sin x cos x

Combine the derivatives of both terms:

-6 sin 2x + 2 sin x cos x

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Answer from HIX Tutor

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.

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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