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,

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

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