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

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Answer 1 · 2016-01-04T09:59:28

#dy/dx=3x^2#

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#