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How do you find the derivative of #g(x)= 3tan4xsin2xcos2x#?

Answer 1

Charles Anctil

First just remember

#sin(a)*cos(a)=1/2sin(2a)#

#3/2tan(4x)sin(4x)#

#=3/2sin^2(4x)/cos(4x)#

Remember #sin^2(a) = 1-cos^2(a)#

#3/(2cos(4x))-3/2cos(4x)#

#(3/2cos^-1(4x))' = 3/2(cos^-1(4x))' = 6sin(4x)*cos^-2(4x)#

#(u^n)'=n*u'*u^(n-1)#

#(3/2cos(4x))'=3/2(cos(4x))' = -6sin(4x)#

So the result is

#=6sin(4x)*cos^-2(4x)+6sin(4x)#

#6sin(4x)(cos^-2(4x)+1)#

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Answer 2

Christopher Allers

To find the derivative of ( g(x) = 3\tan(4x)\sin(2x)\cos(2x) ), you can use the product rule and chain rule.

Let's denote:

( u(x) = 3\tan(4x) )

( v(x) = \sin(2x)\cos(2x) )

Apply the product rule:

( g'(x) = u'(x)v(x) + u(x)v'(x) )

Now, differentiate ( u(x) ) and ( v(x) ) separately using the chain rule and product rule, respectively:

( u'(x) = 3 \cdot \frac{d}{dx}[\tan(4x)] )

( v'(x) = \frac{d}{dx}[\sin(2x)]\cos(2x) + \sin(2x)\frac{d}{dx}[\cos(2x)] )

After finding the derivatives, substitute back into the product rule formula and simplify to find ( g'(x) ).

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Answer from HIX Tutor

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