Find the second derivative.
f(x)=(2x^2+4)^7/2
I assume you meant
f(x)=(2x^2+4)^(7/2), since the "/2" at the end is kind of weird otherwise.
So, use the power rule and chain rule. If
f = u^n, f' = n u^(n-1) u'
f' = (7/2)(2x^2+4)^(5/2) (4x)
= 14x (2x^2+4)^(5/2)
how do u find the second derivative?
To find the second derivative of a function, we need to find the derivative of its first derivative. Let's start by finding the first derivative of f(x).
The function f(x) is given as f(x) = (2x^2 + 4)^(7/2).
To find the derivative, we will use the chain rule. The chain rule states that if we have a composite function, f(g(x)), then the derivative is given by f'(g(x)) * g'(x).
First, we need to find the derivative of the inner function, 2x^2 + 4. The derivative of this function is:
d/dx (2x^2 + 4) = 4x
Next, we differentiate the outer function, (2x^2 + 4)^(7/2).
To do this, we can rewrite the function as (u)^n, where u = (2x^2 + 4) and n = 7/2.
Using the power rule, the derivative of (u)^n is given by n(u)^(n-1) * du/dx.
So, the derivative of (2x^2 + 4)^(7/2) is:
(7/2) * (2x^2 + 4)^(7/2 - 1) * (d/dx (2x^2 + 4))
= (7/2) * (2x^2 + 4)^(5/2) * (4x)
Simplifying further:
= 14x * (2x^2 + 4)^(5/2)
Now, to find the second derivative, we differentiate the first derivative obtained above.
Taking the derivative of 14x * (2x^2 + 4)^(5/2):
= 14 * (2x^2 + 4)^(5/2) + 14x * (5/2) * (2x^2 + 4)^(5/2 - 1) * (d/dx (2x^2 + 4))
= 14 * (2x^2 + 4)^(5/2) + (35/2) * x * (2x^2 + 4)^(3/2) * (4x)
= 14 * (2x^2 + 4)^(5/2) + 70x^2 * (2x^2 + 4)^(3/2)
Therefore, the second derivative of f(x) = (2x^2+4)^(7/2) is:
f''(x) = 14 * (2x^2 + 4)^(5/2) + 70x^2 * (2x^2 + 4)^(3/2)