How do you find the derivative for #y= (x^2 + 2x + 5) / (x + 1)#?

Question

Christopher Allers · Accepted Answer

#y^' = 1 - 4/(x+1)^2#

You could differentiate this function by using the quotient rule, or you could simplify the function first, then use the chain rule.

Here's how you could simplify this function. Notice that the numerator can be written as

#y = (x^2 + 2x + 1 + 4)/(x+1)#

This is an important thing to notice because you can use the formula for the square of a binomial

#color(blue)( (a+b)^2 = a^2 + 2ab + b^2)#

to write

#y = ((x + 1)^2 + 4)/(x+1)#

You can then go ahead and simplify this by

#y = (x+1)^color(red)(cancel(color(black)(2)))/(color(red)(cancel(color(black)((x+1))))) + 4/(x+1)#

#y = x + 1 + 4/(x+1)#This can be differentiated by using the chain rule for

#d/dx((x+1)^(-1)) = d/(du)u^(-1) * d/dx(u)#

#d/dx((x+1)^(-1)) = -u^(-2) * d/dx(x+1)#

#d/dx((x+1)^(-1)) = -(x+1)^(-2)#

Your target derivative will thus be equal to

#y^' = d/dx(x) + d/dx(1) + 4 * [-(x+1)^(-2)]#

#y^' = 1 + 0 - 4 * 1/(x+1)^2 = color(green)(1-4/(x+1)^2)#

Isaiah Allen · Answer

undefined To find the derivative of y = (x^2 + 2x + 5) / (x + 1), we'll use the quotient rule. The quotient rule states that if you have a function in the form of f(x)/g(x), then the derivative is (g(x) * f'(x) - f(x) * g'(x)) / (g(x))^2. Applying this to the given function:

f(x) = x^2 + 2x + 5
g(x) = x + 1

Now, we'll find the derivatives of f(x) and g(x):

f'(x) = 2x + 2
g'(x) = 1

Now, applying the quotient rule:

y' = [(x + 1)(2x + 2) - (x^2 + 2x + 5)(1)] / (x + 1)^2

Expanding and simplifying:

y' = [2x^2 + 2x + 2x + 2 - x^2 - 2x - 5] / (x + 1)^2
y' = (x^2 + 4x - 3) / (x + 1)^2

So, the derivative of y = (x^2 + 2x + 5) / (x + 1) is y' = (x^2 + 4x - 3) / (x + 1)^2.