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How do you differentiate #f(x)=2sinxcosx#?

Answer 1

Emilia Aguilar

#f'(x) = 2cos(2x)#

Use the identity #2sinthetacostheta = sin2theta# to rewrite the function.

#f(x) = sin2x#

Now, use the chain rule to differentiate.

Let #y = sinu# and #u = 2x#. #y' = cosu# and #u' = 2#.

#dy/dx = cosu xx 2 = cos(2x) xx 2 = 2cos(2x)#

Hopefully this helps!

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Answer 2

Elijah Abram

To differentiate ( f(x) = 2 \sin(x) \cos(x) ), you can use the product rule of differentiation. The product rule states that if you have two functions, ( u(x) ) and ( v(x) ), then the derivative of their product is given by ( \frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x) ). Applying this rule to ( f(x) = 2 \sin(x) \cos(x) ), where ( u(x) = 2 \sin(x) ) and ( v(x) = \cos(x) ), we have:

[ f'(x) = (2 \sin(x))' \cdot \cos(x) + 2 \sin(x) \cdot (\cos(x))' ]

[ f'(x) = (2 \cos(x)) \cdot \cos(x) + 2 \sin(x) \cdot (-\sin(x)) ]

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

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

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Answer from HIX Tutor

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