Asked by Angie
I'm having problems with this one. Can't get the right answer.
Differentiate the following using the chain rule.
f(x) = squareroot(x^2+1)/(3x+1)
I know that I can take the whole thing and put it to the (1/2) and differentiate that then use the quotient rule to differentiate the inside, but when i try to factor I can't get stuck. with 3x^2+2x-3/(3x+1)^2
Differentiate the following using the chain rule.
f(x) = squareroot(x^2+1)/(3x+1)
I know that I can take the whole thing and put it to the (1/2) and differentiate that then use the quotient rule to differentiate the inside, but when i try to factor I can't get stuck. with 3x^2+2x-3/(3x+1)^2
Answers
Answered by
Steve
assuming you mean
f(x) = √((x^2+1)/(3x+1))
f = u^1/2 where u = (x^2+1)/(3x+1)
so,
f' = 1/2√u u'
u = f/g where
f = x^2+1 and v = 3x+1
so, u' = (f'g-g'f)/g^2 = (3x^2+2x-3)/(3x+1)^2
so f' = 1/[2√((x^2+1)/(3x+1))] * (3x^2+2x-3)/(3x+1)^2
= (3x^2+2x-3)/[2(3x+1)<sup>3/2</sup>(x^2+1)<sup>1/2</sup>]
f(x) = √((x^2+1)/(3x+1))
f = u^1/2 where u = (x^2+1)/(3x+1)
so,
f' = 1/2√u u'
u = f/g where
f = x^2+1 and v = 3x+1
so, u' = (f'g-g'f)/g^2 = (3x^2+2x-3)/(3x+1)^2
so f' = 1/[2√((x^2+1)/(3x+1))] * (3x^2+2x-3)/(3x+1)^2
= (3x^2+2x-3)/[2(3x+1)<sup>3/2</sup>(x^2+1)<sup>1/2</sup>]
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