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HELP! Derivative problems, involving the quotient rule and trig!?
The question is:
Given f(x) = 2x/sin3x, find f' of (pi/3).
My work is as follows:
f'(x) = ((sin3x)(2) - 2x(3cos3x))/(sin3x)2
However, this isn't right! Where did I go wrong? And the denominator is equal to zero! How does this work?
The question is:
Given f(x) = 2x/sin3x, find f' of (pi/3).
My work is as follows:
f'(x) = ((sin3x)(2) - 2x(3cos3x))/(sin3x)^2
However, this isn't right! Where did I go wrong? And the denominator is equal to zero! How does this work?
3 Answers
- 1 decade agoFavorite Answer
think you are close but,
your first term should be 2/sin3x,
just the d/dx(2x)/sin3x
I think you've got then the second term right but just left of the power
f'(x) = 2/(sin3x) - 2x(3cos3x)/(sin3x)^2
- 1 decade ago
I just learned this, you don't go to Elder do you?
would it would be f'= ((sin3x)(2)- (2x)(cos3))/(sin3x)2
- Anonymous4 years ago
for this issue you wouldnt favor chain rule. you take advantage of chain rule once you've a function interior of a function. so as an example, sin(x^2 + a million). you wanna artwork outdoors to interior so that you may want to first commence with the sin element. by-product of sin is cos. so that you may want to get cos(x^2 + a million). now flow on to the interior. the by-product of x^2 + a million is an similar as 2x. so now multiply those mutually and also you get 2xcos(x^2 + a million). thats your answer.
