Question

Find the derivative of the algebraic functions

Answer 1

Answer:

10) f'(x) = (x² + 6x - 3)/(x + 3)²

11) g'(x) = 1 - csc²x

Step-by-step explanation:

10. f(x) = x(1 - (4/(x + 3))

Expanding gives;

f(x) = x - (4x/(x + 3))

Differentiating gives;

f'(x) = 1 - 4/(x + 3) + 4x/(x + 3)²

Simplifying this gives;

f'(x) = [(x + 3)² - 4(x + 3) + 4x]/(x + 3)²

f'(x) = (x² + 6x + 9 - 4x - 12 + 4x)/(x + 3)²

f'(x) = (x² + 6x - 3)/(x + 3)²

11. g(x) = x + cot x

Rewriting this gives;

g(x) = x + (1/tan x)

We know that derivative of tan x is sec x while derivative of (1/tan x) is -csc²x

Thus;

g'(x) = 1 - csc²x

This can be written as

Differentiating this gives;

g'(x) = 1 - csc²x