Question 560133
This is a calculus 1 question.


You need to find the derivative of F(x).


I'm looking for F'(x).  So, I will do this by differenting termwise.


(4x^3)' = 12x^2


(-5x)' = -5


(2x^1/2)' = x^(-1/2)


We can write  x^(-1/2) as 1/sqrt{x}.


So, F'(x) = 12x^2 + 1/sqrt{x} - 5


To find the slope of the tangent line, we now replace every x you see with 4 and simplify.


F'(4) = 12(4)^2 + 1/sqrt{4} - 5


F'(4) = 12*16 + 1/2 - 5


F'(4) = 375/2


The slope of the tangent line is 375/2.