How do you find the derivative for #(x^2 + 3) / x#?

3 Answers
Jul 31, 2015

I found: #(x^2-3)/x^2#

Explanation:

You can use the Quotient Rule where the derivative of #f(x)/g(x)# is:
#(f'(x)g(x)-f(x)g'(x))/[g(x)]^2#
in your case you get:

#(2x*x-(x^2+3)*1)/x^2=(2x^2-x^2-3)/x^2=(x^2-3)/x^2#

Jul 31, 2015

Alternatively, you could differentiate this function by using the product rule.

Explanation:

You could rewrite your original function as

#y = (x^2 + 3) * x^(-1#

This will allow you to use the product rule, which tells you that functions that can be expressed as the product of two other functions

#y = f(x) * g(x)#

can be differentiate by

#color(blue)(d/dx(y) = f^'(x)g(x) + f(x)g^'(x)#

In your case, you have

#f(x) = x^2 + 3# and #g(x) = x^(-1)#, so that your derivative will be

#d/dx(y) = [d/dx(x^2 + 3)] * x^(-1) + (x^2 + 3) * d/dx(x^(-1))#

#d/dx(y) = 2 * color(red)(cancel(color(black)(x))) * 1/color(red)(cancel(color(black)(x))) + (x^2 + 3) * (-x^(-2))#

#d/dx(y) = 2 - (x^2 + 3)/x^2#

#d/dx(y) = 2 - color(red)(cancel(color(black)(x^2)))/color(red)(cancel(color(black)(x^2))) - 3/x^2 = color(green)(1 - 3/x^2)#

Jul 31, 2015

Alternatively again, you could use the power rule.

Explanation:

#f(x)=(x^2+3)/(x)#

#=x^2/x+3/x#

#=x+3/x#

#=x+3x^(-1)#

#f'(x)=1-3x^(-2)#

#f'(x)=1-3/x^(2)#