find the derivative of
y= x^3+8/x+2
the / is meant to be a fraction bar, I am really really bad at finding the derivative of a fraction function, could you help by showing me exactly how, Thanks
Do you mean?
y= (x^3+8)/(x+2) ???? or what you typed?
If my guess is correct then
(bottom dtop/dx -top d bottom/dx)/bottom^2
=[(x+2)(3x^2)-(x^3+3)(1)]/(x+2)^2
=[3x^3+6x^2-x^3-3]/(x+2)^2
= (2x^3+6x^2-3)/(x+2)^2
Thanks
You are welcome.
Of course! I'll be happy to help you find the derivative of the given function, which involves both the power rule and the quotient rule. Let's break it down step by step.
The function you provided is:
y = (x^3 + 8)/(x + 2)
To find the derivative, we'll use the quotient rule, which states that for any two functions, u(x) and v(x):
[d(u(x)/v(x))/dx] = (v(x) * d(u(x))/dx - u(x) * d(v(x))/dx) / (v(x))^2
Let's apply this rule to our function, where u(x) = (x^3 + 8) and v(x) = (x + 2):
Step 1: Find d(u(x))/dx (the derivative of the numerator).
To differentiate the numerator, we need to use the power rule, which states that for any function of the form f(x) = x^n:
d/dx (x^n) = n * x^(n-1).
So, applying the power rule to u(x) = (x^3 + 8), we get:
d(u(x))/dx = 3x^(3-1) = 3x^2.
Step 2: Find d(v(x))/dx (the derivative of the denominator).
The derivative of the denominator v(x) = (x + 2) is simply 1 since it is a linear term.
d(v(x))/dx = 1.
Step 3: Apply the quotient rule to the numerator and denominator derivatives.
Substituting the derivatives we found into the quotient rule formula, we get:
[d(y)/dx] = [(x + 2) * (3x^2) - (x^3 + 8) * (1)] / (x + 2)^2
Simplifying this expression, we have:
[d(y)/dx] = (3x^3 + 6x^2 - x^3 - 8) / (x + 2)^2
Combining like terms in the numerator, we get:
[d(y)/dx] = (2x^3 + 6x^2 - 8) / (x + 2)^2
So, the derivative of y with respect to x can be written as:
dy/dx = (2x^3 + 6x^2 - 8) / (x + 2)^2.
That's it! We've successfully found the derivative of the given function using the quotient rule.