This is a large problem, sorry for the trouble.
if f(x) = sqrt(x^3 + 5x + 121)(x^2 + x + 11)
Find fprime/derivative of/at 0
I keep getting 16 which is not one of the multiple choice answers available.
My last step is [sqrt(121)(1)] + [{5/sqrt(121)}(11)]
What am I doing wrong?
To find the derivative of the function f(x), you can use the product rule. The product rule states that if you have two functions, say u(x) and v(x), their derivative can be found using the formula:
d(uv)/dx = u * dv/dx + v * du/dx
In this case, u(x) = sqrt(x^3 + 5x + 121) and v(x) = x^2 + x + 11. Let's find the derivatives of u(x) and v(x) first.
To find du/dx, you can use the chain rule. The chain rule states that if you have a composition of functions, say u(g(x)), then the derivative of that composition can be found using the formula:
du/dx = du/dg * dg/dx
In this case, u(x) = sqrt(x^3 + 5x + 121), and g(x) = x^3 + 5x + 121. Let's find the derivative of g(x).
dg/dx = 3x^2 + 5
Now let's find du/dg.
du/dg = 1/(2 * sqrt(g(x)))
Substituting the value of g(x) into du/dg gives us:
du/dg = 1/(2 * sqrt(x^3 + 5x + 121))
Now let's find the derivative of v(x).
dv/dx = 2x + 1
Now, using the product rule, we can find the derivative of f(x).
f'(x) = u * dv/dx + v * du/dx
Substituting the values:
f'(x) = sqrt(x^3 + 5x + 121) * (2x + 1) + (x^2 + x + 11) * (1/(2 * sqrt(x^3 + 5x + 121)))
To find the derivative at x = 0, you need to substitute x = 0 into the derivative expression:
f'(0) = sqrt((0)^3 + 5(0) + 121) * (2(0) + 1) + ((0)^2 + (0) + 11) * (1/(2 * sqrt((0)^3 + 5(0) + 121)))
Simplifying this expression gives us:
f'(0) = sqrt(121) * (0 + 1) + 11 * (1/(2 * sqrt(121)))
Simplifying further gives us:
f'(0) = 11 * (1/(2 * 11)) = 1/2
So the derivative of f(x) at x = 0 is 1/2.
To find the derivative of f(x) at x = 0, you need to find the derivative of each term individually using the product rule.
Let's break down the function f(x) = sqrt(x^3 + 5x + 121)(x^2 + x + 11):
Step 1: Apply the product rule to find the derivative of the first term.
Let g(x) = sqrt(x^3 + 5x + 121) and h(x) = x^2 + x + 11.
Using the product rule, we have:
g'(x) = (1/2√(x^3 + 5x + 121)) * (3x^2 + 5)
h'(x) = 2x + 1
Now, we can find the derivative of the first term by using the product rule:
f1'(x) = g(x) * h'(x) + g'(x) * h(x)
= sqrt(x^3 + 5x + 121) * (2x + 1) + (1/2√(x^3 + 5x + 121)) * (3x^2 + 5) * (x^2 + x + 11)
Step 2: Apply the product rule to find the derivative of the second term.
Let j(x) = x^2 + x + 11 and k(x) = sqrt(x^3 + 5x + 121).
Using the product rule, we have:
j'(x) = 2x + 1
k'(x) = (1/2√(x^3 + 5x + 121)) * (3x^2 + 5)
Now, we can find the derivative of the second term by using the product rule:
f2'(x) = j(x) * k'(x) + j'(x) * k(x)
= (x^2 + x + 11) * (1/2√(x^3 + 5x + 121)) * (3x^2 + 5) + (2x + 1) * sqrt(x^3 + 5x + 121)
Step 3: Finally, we can find the derivative of f(x) by adding the derivatives of the individual terms:
f'(x) = f1'(x) + f2'(x)
To find fprime at x = 0, evaluate f'(x) at x = 0:
fprime(0) = f'(0) = f1'(0) + f2'(0)
Substitute x = 0 into the derived functions f1'(x) and f2'(x):
fprime(0) = f1'(0) + f2'(0)
Now, you can calculate fprime(0) by plugging in x = 0 into the derived expressions for f1'(x) and f2'(x) and simplifying the result.