Answer:
10) f'(x) = (x² + 6x - 3)/(x + 3)²
11) g'(x) = 1 - csc²x
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