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Tony2001 Tony2001
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11 years ago
I'm confused as to how to apply the chain rule in this case...

I have to find the derivative of a quotient but there are two terms multiplyed by each other in the numerator and the denominator.

y= (4x)(sin(x)) / (3x-1)(x-1)^1/2

How do I go about starting this question - do I first apply the quotient rule?
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wrote...
11 years ago
use the quotient rule first and then the other rules...
answer; 4x cos(x)/(sqrt(x-1)(3x-1))-(2(3x^2+x-2) sin(x)/((x-1)^(3/2) (3x-1))
wrote...
11 years ago
I think it would be a lot easier if you rewrite the equation:
y=
(4x)(sin(x))*((3x-1)^-1)*((x-1)^-1/2)
and then use the product rule.  I always find
quotient rule is harder to remember...
wrote...
11 years ago
Yes start withthe quotient rule.  You really don't need to use the chain rule here.  

The chain rule is needed when dealing with functions of one or more variables which are themselves functions of other variables, such as e^sin(x) where f(x) = e^x and g(x)=sin(x) and you want to find f'(g(x)).

Here you can just substitute

u=4x, v=sin(x) w=(3x-1) z=?(x-1)

y=uv/(wz)

Since every function is of the same variable the chain rule doesn't apply.  Apply the quotient rule instead:

y' = [(wz) (uv)' - (uv)(wz)']/(wz)²

Now apply the product rule

(uv)' = u'v + uv' and (wz)' = w'z + wz'

y' = [(wz) (u'v + uv') - (uv)(w'z + wz')]/(wz)²

Susbstitute the following back in

u=4x, v=sin(x) w=(3x-1) z=?(x-1)
u'=4, v'=cos(x), w'=3, z'=1/[2?(x-1)]

y' = [(3x-1)?(x-1) (4sin(x) + 4cos(x)) -
...... (4xsin(x))(3?(x-1) + (3x-1)/(2?(x-1))]/[(3x-1)?(x-1)]²

It's quite ugly.  Just use algebra to simplify it.
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