Answer :
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].