How do you differentiate #g(x) = sin(3x)sin(6x)# using the product rule?
According to the product rule, if
then
In the instance above,
so
so
Consequently, if you recall the product rule equation, the derivative of the entire thing is
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To differentiate ( g(x) = \sin(3x)\sin(6x) ) using the product rule, you apply the formula ( (uv)' = u'v + uv' ). Let ( u = \sin(3x) ) and ( v = \sin(6x) ). The derivatives are ( u' = 3\cos(3x) ) and ( v' = 6\cos(6x) ). Therefore, the derivative of ( g(x) ) is ( g'(x) = (3\cos(3x))(\sin(6x)) + (\sin(3x))(6\cos(6x)) ).
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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.
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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