#3**+1 **

The value of f' at x = 2 = slope of the curve at that point....

slope of tangent line 1 = 4/2 slope of line 2 = - 1/3

slope 1 * slope 2 = -4/6 = -2/3

ElectricPavlov Jan 11, 2019

#6**+1 **

In both diagrams, the tangent line is given. By definition of derivative, the derivatives of the functions at the point of tangency is just the slope of tangent line.

Therefore, by calculating(or approximating? :D) the slope of the tangents, we get \(p'(2) = 2\) and \(q'(2) = \dfrac{-1}{3}\).

And then, we see the graph of p(x) passes through (2,3), and the graph of q(x) passes through (2, -1). So p(2) = 3 and q(2) = -1.

By product rule of differentiation:

\(\quad r'(2)\\ = p(2) q'(2) + q(2) p'(2)\\ = 3\cdot \dfrac{-1}{3} + (-1)\cdot 2\\ = -1 - 2\\ = -3\)

There we go :)

MaxWong Jan 14, 2019