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
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)3/2(x^2+1)1/2]
To differentiate the function f(x) = √(x^2+1)/(3x+1) using the chain rule, you are on the right track.
First, let's denote the numerator as u(x) = √(x^2+1) and the denominator as v(x) = 3x+1.
Now, let's find u'(x), which is the derivative of the numerator:
u'(x) = d/dx (√(x^2+1))
To differentiate this, you can use the chain rule, where you differentiate the outer function (√), multiply it by the derivative of the inner function (x^2+1).
The derivative of √(x^2+1) can be found using the power rule. Let's call √(x^2+1) as w(x):
w'(x) = d/dx (√(x^2+1))
= (1/2) * (x^2+1)^(-1/2) * d/dx (x^2+1)
= (1/2) * (x^2+1)^(-1/2) * 2x
= x / √(x^2+1)
So, u'(x) = x / √(x^2+1).
Next, let's find v'(x), which is the derivative of the denominator v(x) = 3x+1:
v'(x) = d/dx (3x+1)
= 3
Now, let's use the quotient rule to differentiate the function f(x) = u(x)/v(x):
f'(x) = (v(x) * u'(x) - u(x) * v'(x)) / v(x)^2
Substituting the values, we get:
f'(x) = [(3x+1) * (x / √(x^2+1)) - √(x^2+1) * 3] / (3x+1)^2
= [3x^2 + x - 3√(x^2+1)] / (3x+1)^2
So, the derivative of f(x) is [3x^2 + x - 3√(x^2+1)] / (3x+1)^2.
Remember to simplify further if required.