Find the value of (f o g)' at the given value of x.
f(u)= u/u^2-1 u=g(x)=3x^2+x+3 x=0
is it a. 5/32 b. 13/32 c. 1/8 d. 5/32
f(u) = u/(u^2-1)
df/du = [(u^2-1) - 2u^2]/(u^2-1)^2
= - (u^2+1)/(u^2-1)^2
fog is just f(g(x)) = f(u)
so, (fog)' is just d/dx(f(g(x)) = df/dx
df/dx = df/du * du/dx =
-(u^2+1)/(u^2-1)^2 * (6x+1)
u(0) = 3
f'(3) = -10/8^2 * 1 = -5/32
I don't see that as one of the choices, but I do see that (a) and (d) are both 5/32, so I assume one of them was -5/32
To find the value of (f o g)' at the given value of x, we need to find the derivative of f(g(x)) with respect to x and evaluate it at x=0.
Step 1: Find f'(u)
Let's begin by finding the derivative of f(u).
f(u) = u / (u^2 - 1)
Using the quotient rule, we can differentiate f(u) with respect to u:
f'(u) = (u^2 - 1)(1) - u(2u) / (u^2 - 1)^2
= (u^2 - 1 - 2u^2) / (u^2 - 1)^2
= (-u^2 - 1) / (u^2 - 1)^2
Step 2: Find g'(x)
Next, we need to find the derivative of g(x).
g(x) = 3x^2 + x + 3
To find g'(x), we differentiate each term with respect to x:
g'(x) = (d/dx) (3x^2) + (d/dx) (x) + (d/dx) (3)
= 6x + 1
Step 3: Substitute g'(x) into f'(u)
Now, we substitute g'(x) into f'(u):
f'(g(x)) = (-g(x)^2 - 1) / (g(x)^2 - 1)^2
= (-(3x^2 + x + 3)^2 - 1) / ((3x^2 + x + 3)^2 - 1)^2
Step 4: Evaluate at x=0
Finally, we substitute x=0 into f'(g(x)):
(f o g)'(0) = (-(3(0)^2 + 0 + 3)^2 - 1) / ((3(0)^2 + 0 + 3)^2 - 1)^2
Simplifying:
(f o g)'(0) = (-(3)^2 - 1) / (3^2 - 1)^2
= (-9 - 1) / (9 - 1)^2
= (-10) / 8^2
= -10 / 64
= -5 / 32
Therefore, the value of (f o g)' at x=0 is -5/32.
The correct answer is d. -5/32.
To find the value of (f o g)' at a given value of x, we need to find the derivative of the composite function f(g(x)) and evaluate it at the given value of x.
First, let's find f'(u), the derivative of f(u) with respect to u:
f(u) = u / (u^2 - 1)
To find f'(u), we can apply the quotient rule:
f'(u) = (u^2 - 1)(1) - u(2u) / (u^2 - 1)^2
= (u^2 - 1 - 2u^2) / (u^2 - 1)^2
= (-u^2 - 1) / (u^2 - 1)^2
Next, let's find g'(x), the derivative of g(x) with respect to x:
g(x) = 3x^2 + x + 3
To find g'(x), we can simply differentiate each term:
g'(x) = d(3x^2)/dx + d(x)/dx + d(3)/dx
= 6x + 1 + 0
= 6x + 1
Now, we can find the derivative of the composite function f(g(x)):
(f o g)'(x) = f'(g(x)) * g'(x)
Substituting f'(u) and g'(x) into the equation, we have:
(f o g)'(x) = [(-g(x)^2 - 1) / (g(x)^2 - 1)^2] * (6x + 1)
Substituting g(x) = 3x^2 + x + 3, we have:
(f o g)'(x) = [(-(3x^2 + x + 3)^2 - 1) / ((3x^2 + x + 3)^2 - 1)^2] * (6x + 1)
Finally, to find the value of (f o g)' at x = 0, substitute x = 0 into the equation:
(f o g)'(0) = [(-(3(0)^2 + 0 + 3)^2 - 1) / ((3(0)^2 + 0 + 3)^2 - 1)^2] * (6(0) + 1)
= [(-(3)^2 - 1) / ((3)^2 - 1)^2] * 1
= [(-(9) - 1) / (9 - 1)^2] * 1
= (-10 / 64) * 1
= -10 / 64
= -5 / 32
Therefore, the value of (f o g)' at x = 0 is -5/32.
The correct answer is not among the options provided.