Help with inside/outside functions using the chain rule?

I don't know what "inside/outside" means here. I think of it as "unwrapping" the function so that's probably related to your terminology.

(5x^4+3x)^3 is an expression cubed. So let's write it as f(x)^3.
Then it should be obvious that the derivative is 3 f(x)^2 * f'(x) by the chain rule. And since f(x) = 5x^4 + 3x, it's a simple polynomial and you don't need any other rules to break it down.

sqrt[ sqrt(5x^2 + 3x) ] means (5x^2+3x)^(1/4) and you can apply exactly the same logic as above. This is in the form f(x)^(1/4) and the derivative is (1/4) f(x)^(-3/4) * f'(x)

5/f(x)^2 is the same as 5 f(x)^(-2) and by the EXACT SAME REASONING, the derivative is 5 * (-2) f(x)^(-3) * f'(x)

All of these are of the form f(x)^n for some power n and the derivative is n f(x)^(n-1) * f'(x)....Show more

1. the cubing function f is outside, the function g which takes x to 5x^4+3x is inside and the given function takes x to f(g(x)).

2. the square root function is outside, the function x-->5x^2+3x is inside
(I assume you mean sqrt(5x^2+3x), but you left off parentheses leading to confusion)

3. more than one way will work ... outside function x-->5/x^2, inside function x-->(1/2)x^3 - x^-2.

Note: a+b / c+d is NOT the same as (a+b) / (c+d) and √a+b is NOT the same as √(a+b)...Show more