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How do you differentiate #f(x) =sin(3x-2)* cos(3x -2) #?

Answer 1

Charles Arocha

Remember that

#sin2A=2*sinA*cosA#

Hence

#f(x)=sin(3x-2)*cos(3x-2)=>f(x)=1/2*sin(2*(3x-2))#

Hence

#(df)/dx=1/2*cos(2(3x-2))*6=3*cos(6x-4)=#

Finally

#f'(x)=3*cos(6x-4)#

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Answer 2

Paisley Johnson

To differentiate ( f(x) = \sin(3x-2) \cdot \cos(3x-2) ), you can use the product rule, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. Applying this rule:

( f'(x) = \frac{d}{dx}(\sin(3x-2)) \cdot \cos(3x-2) + \sin(3x-2) \cdot \frac{d}{dx}(\cos(3x-2)) )

Now, differentiate each part:

( \frac{d}{dx}(\sin(3x-2)) = 3 \cos(3x-2) )

( \frac{d}{dx}(\cos(3x-2)) = -3 \sin(3x-2) )

Substitute these derivatives back into the formula:

( f'(x) = 3 \cos(3x-2) \cdot \cos(3x-2) - 3 \sin(3x-2) \cdot \sin(3x-2) )

( f'(x) = 3 \cos^2(3x-2) - 3 \sin^2(3x-2) )

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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.