Need help with the following along with all steps.
Find y" for y = 3x + 4/x
write it as
y = 3x + 4x^-1
y' = 3 - 4x^-2
y'' = 8x^-3 or 8/x^3
Thank you so much for your help. Now that I look at it more closely. I understand what I wasn't seeing!
To find the second derivative, we need to differentiate the function twice. Let's start by finding the first derivative of y with respect to x.
Step 1: Begin by differentiating each term separately.
- The derivative of 3x is 3.
- The derivative of 4/x can be found using the quotient rule.
Step 2: Apply the quotient rule to find the derivative of 4/x.
- The quotient rule states that the derivative of f(x)/g(x) is given by:
(g(x)f'(x) - f(x)g'(x))/ [g(x)]^2.
- In this case, f(x) = 4 and g(x) = x.
Step 3: Differentiate f(x) and g(x) accordingly.
- The derivative of f(x) = 4 is 0 (since it is a constant).
- The derivative of g(x) = x is 1.
Step 4: Apply the quotient rule to find the derivative.
- (g(x)f'(x) - f(x)g'(x))/ [g(x)]^2 = (1*0 - 4*1)/(x^2) = -4/x^2.
Step 5: Now, we have found the first derivative of y.
- y' = 3 - 4/x^2.
To find the second derivative (y"), we follow the same process:
Step 6: Differentiate the first derivative (y') with respect to x.
- Differentiating 3 with respect to x gives us 0 (since it is a constant).
- To differentiate -4/x^2, we use the quotient rule again.
Step 7: Apply the quotient rule to find the second derivative.
- (g(x)f'(x) - f(x)g'(x))/ [g(x)]^2 = (1*8/x^3 - (-4/x^2)*1)/(x^2) = (8/x^3 + 4/x^3)/(x^2) = 12/x^5.
Step 8: Simplify the expression.
- y" = 12/x^5.
Therefore, the second derivative of y = 3x + 4/x is y" = 12/x^5.