calculus
posted by Adam Janky .
Consider the following.
y = (2x2 + 5)(x3 − 25x)
at (5, 0)
(a) At the indicated point, find the slope of the tangent line.
(b) Find the instantaneous rate of change of the function.

First check if (5,0) is on the line:
f(x)= (2x2 + 5)(x3 − 25x)
f(5)= (50+5)(125125)=0 indeed.
a)Slope
Slope of the tangent line at (5,0) is the value of f'(x) at (5,0), namely f'(5).
So let's find f'(x).
We can find f'(x) as a product, or as an expanded polynomial. I choose the latter:
y = (2x2 + 5)(x3 − 25x)
=2x^5+5x^350x^3125x^2
=2x^545x^3125x^2
Differentiate term by term:
f'(x)=10x^4135x2250x
so
f'(5)=625033751250=1625
= slope of tangent at (5,0)
b.
instantaneous rate of change of the function is precisely f'(x)=dy/dx.
So the answer is the same as in a.