What is the slope of the tangent of f(x) at the point =1 ?

$f(x)=(x^2-4)^3$

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

$f'(1)=6.(1).(1^2-4)^2=6(1).(-3)^2=54$

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