Given f, a derivable function and
y = f (x)^3 + 40√(f(x)+12) .
Knowing that f(4) = -4
and that dy/dx|x=4 = 14 ,
calculate f’(4)
To find f'(4), we can use the chain rule of differentiation. The chain rule states that if we have a composite function, we can differentiate it by multiplying the derivative of the outer function with the derivative of the inner function.
In this case, we have y = f(x)^3 + 40√(f(x)+12). Let's break it down and differentiate each part separately.
First, let's differentiate the term f(x)^3. This is a power rule, where the derivative of f(x)^n is n*f(x)^(n-1). In this case, n = 3, so the derivative is 3*f(x)^2.
Next, let's differentiate the term 40√(f(x)+12). This involves the chain rule. The derivative of √(u) is (1/2) * (1/√(u)) * du/dx, where u is the inner function. In this case, the inner function is f(x) + 12. So, the derivative is (1/2) * (1/√(f(x)+12)) * (d(f(x)+12)/dx).
Now, let's differentiate the inner function (f(x) + 12). Since f(4) = -4, we know that f(x) = -4 when x = 4. So, f'(4) is the derivative of f(x) at x = 4.
Finally, let's put everything together. We have y = f(x)^3 + 40√(f(x)+12), and we want to find f'(4).
Taking the derivative of y with respect to x, we get:
dy/dx = 3*f(x)^2 * f'(x) + (1/2) * (1/√(f(x)+12)) * (d(f(x)+12)/dx)
Substituting x = 4, we have:
14 = 3*f(4)^2 * f'(4) + (1/2) * (1/√(f(4)+12)) * (d(f(4)+12)/dx)
We know that f(4) = -4, so we can substitute that value in:
14 = 3*(-4)^2 * f'(4) + (1/2) * (1/√(-4+12)) * (d(-4+12)/dx)
Simplifying further:
14 = 3*16 * f'(4) + (1/2) * (1/√8) * (0)
Since d(-4+12)/dx is 0, we can disregard that term.
Now, we can solve the equation for f'(4):
14 = 3*16 * f'(4)
Dividing both sides by 48:
f'(4) = 14/48
Simplifying further, f'(4) = 7/24.
Therefore, f'(4) is equal to 7/24.