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For the curve with equation y=x^3-6x^2+20 , find dy/dx

Sagot :

Given:

The equation of a curve is

[tex]y=x^3-6x^2+20[/tex]

To find:

The values of [tex]\dfrac{dy}{dx}[/tex].

Solution:

Formulae used:

[tex]\dfrac{d}{dx}x^n=nx^{n-1}[/tex]

[tex]\dfrac{d}{dx}C=0[/tex]

We have,

[tex]y=x^3-6x^2+20[/tex]

Differentiate with respect to x.

[tex]\dfrac{dy}{dx}=3x^{3-1}-6(2x^{2-1})+(0)[/tex]

[tex]\dfrac{dy}{dx}=3x^{2}-12x^{1}[/tex]

Therefore, the value of [tex]\dfrac{dy}{dx}[/tex] is [tex]3x^{2}-12x[/tex].

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