Given v(x) = g(x) (3/2*x^4 + 4x – 1), find v'(2).

Step-by-step explanation:

Given that v(x) = g(x)×(3/2*x^4+4x-1)

Let's find V'(2)

V(x) is a product of two functions

● V'(x) = g'(x)×(3/2*x^4+4x-1)+ g(x) ×(3/2*x^4+4x-1)

We are interested in V'(2) so we will replace x by 2 in the expression above.

g'(2) can be deduced from the graph.

● g'(2) is equal to the slope of the tangent line in 2.

● let m be that slope .

● g'(2) = m =>g'(2) = rise /run

● g'(2) = 2/1 =2

We've run 1 square to the right and rised 2 squares up to reach g(2)

g(2) is -1 as shown in the graph.

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Let's derivate the second function.

Let h(x) be that function

● h(x) = 3/2*x^4 +4x-1

● h'(x) = 3/2*4*x^3 + 4

● h'(x) = 6x^3 +4

Let's calculate h'(2)

● h'(2) = 6 × 2^3 + 4

● h'(2) = 52

Let's calculate h(2)

●h(2) = 3/2*2^4 + 4×2 -1

●h(2)= 31

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Replace now everything with its value to find V'(2)

● V'(2) = g'(2)×h(2) + g(2)× h'(2)

● V'(2)= 2×31 + (-1)×52

●V'(2) = 61 -52

●V'(2)= 9