How do you differentiate #y = x^3-8#?

1 Answer
Jan 4, 2016

#dy/dx=3x^2#

Explanation:

To differentiate, you must differentiate every term with #dx#

#dy/dx=(d(x^3))/dx-(d(8))/dx#


Part 1) #(d(x^3))/dx=3x^2#

Note: The formula used is #(d(x^n))/dx=nx^(n-1)#

Since #n=3#,

#(d(x^3))/dx=3x^(3-1)=3x^2#


Part 2) #(d(8))/dx=0#

In this case, #n=0#, since 8 can be expressed as #8x^0#

#(d(8))/dx=(d(8x^0))/dx=8*0x^(0-1)=0#


Therefore,

#dy/dx=3x^2-0=3x^2#