Find the instantaneous rate of change of the surface area s ! 6x2 of a cube with respect to the edge length x at x ! a

The instantaneous rate of change is simply equivalent to the first derivative of the equation or function. We are given the equation of Surface Area (A) with respect to side (x):

A = 6 x^2

Taking the first derivative of the equation:

dA = 12 x dx

dA / dx = 12 x

Now the term dA / dx is the instantaneous rate of change in the surface area with respect to the side length. To get the rate of change when the side x = a, simply plug this in into the equation:

rate of change = dA / dx = 12 x

so when x = a:

rate of change = 12 a

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .