Find the derivative of the function.

Question

g'(x) = ?

Please show all work so I can get a better understanding for this type of problem

Answer 1

Consider this an application of the fundamental theorem of calculus along with the chain rule

Also, note that (u² - 4)/(u² + 4) = 1 - 8/(u² + 4)

g'(x) = (5x)'f(5x) - (3x)' f(3x)= 5(1 - 8/((5x)² + 4) - 3(1 - 8/((3x)² + 4) =

2 + 24/(9x² + 4) - 40/(25x² + 4)

If you wish, this may be written as 2 + (240x² - 64)/(225x⁴ + 136x² + 16)

Answer 2

The fundamental theorem of calculus tells us that

d/dx ∫_a^x f(u) du = f(x)

Thus, if we split up the integral that you are given using the hint that you are given we can say that

d/dx ∫_3x^5x f(u) du = d/dx (∫_3x⁰ f(u) du + ∫₀^5x f(u) du)

d/dx ∫_3x^5x f(u) du = d/dx (-∫₀^3x f(u) du + ∫₀^5x f(u) du)

d/dx ∫_3x^5x f(u) du = d/dx (-∫₀^3x f(u) du) + d/dx (∫₀^5x f(u) du)

d/dx ∫_3x^5x f(u) du = -f(3x) + f(5x)

d/dx ∫_3x^5x f(u) du = -(((3x)² - 4) / ((3x)² + 4)) + (((5x)² - 4) / ((5x)² + 4))

d/dx ∫_3x^5x f(u) du = -((9x² - 4) / (9x² + 4)) + ((25x² - 4) / (25x² + 4))

The typesetting is not very great on here but the idea is you split up the integral, flip the limits on one of them and negate it, and then apply the fundamental theorem of calculus to each integral by distributing the derivative over the integral sum. You then substitute the limits of integration, your function arguments, into the parametric function of u that was being integrated in the first place. The algebraic simplification is left up to you.