Question

how do you find the derivative of h(x)=-x^3+15x^2-48x

Answer 1

Apply (x^n)' = n x^ (n-1)

Answer 2

if y = x^n dy/dx = nx^(n-1)

so h(x) = -x^3 + 15x^2 - 48x

= \[h'(x) = -3\times x^{3-1} + 2\times15x^{2-1} - 4x^{1-1}\]

gives

\[h'(x) = -3x^2 + 30x - 48\]

Answer 3

\[\LARGE h'(x)=(-x^3+15x^2-48x)'\]
\[\LARGE h'(x)=-(x^3)'+(15x^2)'-(48x)'\]
now apply \[\Huge (x^n)'=n\cdot x^{n-1 }\]

Answer 4

Use the power rule.
\[h ^{'}(x)=-3x ^{2}+30x-48\]