Question from NEAP 2007

[lynt.br]

I'm not sure if it's the complete lack of sleep or just my mathematical incompetence but I can't even follow the solutions to this question:

NEAP 2007 MM34 Exam 2 Multiple choice Q22.

Let the function have the rule , where . Let be another continuous function with domain . For all real values of , the derivative of will be equal to:

The answer is

The method in the solutions currently makes no sense to me. If someone could present a simpler, better explained answer than NEAP does, I would much appreciate it.

EDIT: Attached the solutions file if this helps.

[ryley]

To find the derivative of g(f(x)), you need to know the functional notation of the chain rule, which is f'(x)*g'(f(x)). In this case, it simplifies to f'(x)*g'(a^x). Now unlike e^x, you cannot differentiate other numbers raised to x so easily, you have to express them in the form of e^(something) before you can differentiate them. You should know that 'a' can be written as e^(ln(a)), so a^x will equal e(ln(a)*x). This can be differentiated from the rules used in methods, as ln(a) will just be constant, so if f(x)=e(ln(a)*x), f'(x)=ln(a)*e^ln(a)*x), which should be converted back to f'(x)=ln(a)*a^x. Going back to the rule for the derivative of g(f(x)), f'(x)*g'(f(x)) will equal ln(a)*a^x*g'(f(x)), which is a^x*ln(a)*g'(a^x).