How do you differentiate #g(x) = sin(3x)sin(6x)# using the product rule?

Answer 1

#g'(x) = 3sin6xcos3x + 6sin3xcos6x#

According to the product rule, if

#f(x) = a(x) * b(x)#

then

#f'(x) = a'(x)b(x) + b'(x)a(x)#

In the instance above,

#g(x) = sin(3x)sin(6x)#,

so

#a(x) = sin(3x), b(x)=sin(6x)#
Now we have our two separate functions, we can differentiate them. However, these functions have insides (#3x# or #6x#) and outsides (#sin#), so to differentiate them we multiply the derivative of the inside by the derivative of the outside with the inside left the same. In other words,
#d/dx sin(ax + b) = acos(ax + b)#

so

#a(x)=sin(3x), b(x)=sin(6x)#
#a'(x)=3cos(3x), b'(x)=6cos(6x)#

Consequently, if you recall the product rule equation, the derivative of the entire thing is

#g'(x) = 3cos(3x)*sin(6x) + 6cos(6x)*sin(3x)#
#g'(x) = 3sin6xcos3x + 6sin3xcos6x#
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Answer 2

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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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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