Find the rate of change at a point on a polynomial

Mathematica, 6 bytes

#'@#2&

(Beat THAT, MATL and 05AB1E)

The first argument must be a polynomial, with # as its variable and with & at the end (i.e. a pure function polynomial; e.g. 3 #^2 + # - 7 &). The second argument is the x-coordinate of the point of interest.

Explanation

#'

Take the derivative of the first argument (1 is implied).

... @#2&

Plug in the second argument.

Usage

#'@#2&[4 #^3 - 2 #^4 + 5 #^10 &, 19] (* The first test case *)

16134384838410

MATL, 8 6 bytes

yq^**s

Input is: array of exponents, number, array of coefficients.

Try it online! Or verify all test cases: 1, 2 3, 4, 5.

Explanation

Consider example inputs [3 4 10], 19, [4 -2 5].

y    % Take first two inputs implicitly and duplicate the first
     %   STACK: [3 4 10], 19, [3 4 10]
q    % Subtract 1, element-wise
     %   STACK: [3 4 10], 19, [2 3 9]
^    % Power, element-wise
     %   STACK: [3 4 10], [361 6859 322687697779]
*    % Multiply, element-wise
     %   STACK: [1083 27436 3226876977790]
*    % Take third input implicitly and multiply element-wise
     %   STACK: [4332 -54872 16134384888950]
s    % Sum of array
     %   STACK: 16134384838410

Julia, 45 42 40 37 bytes

f(p,x)=sum(i->prod(i)x^abs(i[2]-1),p)

This is a function that acceps a vector of tuples and a number and returns a number. The absolute value is to ensure that the exponent isn't negative, which necessary because Julia annoying throws a DomainError when raising an integer to a negative exponent.

Try it online! (includes all test cases)

Thanks to Glen O for a couple of corrections and bytes.