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

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Answer 1 · 2016-01-04T09:59:28

$\frac{d y}{d x} = 3 x^{2}$ $dy/dx=3x^2$

To differentiate, you must differentiate every term with $d x$ $dx$

$\frac{d y}{d x} = \frac{d (x^{3})}{d x} - \frac{d (8)}{d x}$ $dy/dx=(d(x^3))/dx-(d(8))/dx$

Part 1) $\frac{d (x^{3})}{d x} = 3 x^{2}$ $(d(x^3))/dx=3x^2$

Note: The formula used is $\frac{d (x^{n})}{d x} = n x^{n - 1}$ $(d(x^n))/dx=nx^(n-1)$

Since $n = 3$ $n=3$ ,

$\frac{d (x^{3})}{d x} = 3 x^{3 - 1} = 3 x^{2}$ $(d(x^3))/dx=3x^(3-1)=3x^2$

Part 2) $\frac{d (8)}{d x} = 0$ $(d(8))/dx=0$

In this case, $n = 0$ $n=0$ , since 8 can be expressed as $8 x^{0}$ $8x^0$

$\frac{d (8)}{d x} = \frac{d (8 x^{0})}{d x} = 8 \cdot 0 x^{0 - 1} = 0$ $(d(8))/dx=(d(8x^0))/dx=8*0x^(0-1)=0$

Therefore,

$\frac{d y}{d x} = 3 x^{2} - 0 = 3 x^{2}$ $dy/dx=3x^2-0=3x^2$

How do you differentiate y=x3−8y = x^3-8?