$f(x)=(x^2-4)^3$
What is the slope of the tangent of f(x) at the point=1?
Solution:
Derivative of f(x):
$f'(x)=[(x^2-4)^3]’$
$=3.(x^2-4)^2.(x^2-4)’$
$=3.(x^2-4)^2.2x$
$=6x.(x^2-4)^2$
At point x=1, the slope of the tangent of the function f is
$f'(1)=6.(1).(1^2-4)^2=6(1).(-3)^2=54$
