# calculus

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.

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1. First check if (5,0) is on the line:
f(x)= (2x2 + 5)(x3 − 25x)
f(5)= (50+5)(125-125)=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^3-50x^3-125x^2
=2x^5-45x^3-125x^2
Differentiate term by term:
f'(x)=10x^4-135x2-250x
so
f'(5)=6250-3375-1250=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.

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