How is f'(x)=-x/sqrt(a^2-x^2 when f(x)=a-sqrt(a^2-x^2 using the power rule ?

I learned that the derivative of "x"^n is n"x"^(n-1. So should'nt
d/(dx)[a-sqrt(a^2-x^2)]=-1/(2sqrt(a^2-x^2 where "x"=a^2-x^2 ???

2 Answers
Aug 20, 2017

Ans=-x/sqrt(a^2-x^2)

Explanation:

f(x)=a-sqrt(a^2-x^2)
f'(x)={-1/(2sqrt(a^2-x^2))}{d/dx(a^2-x^2)}
Because here chain rule also be applying
=>-2x/(2sqrt(a^2-x^2))
So the ans is -x/sqrt(a^2-x^2)

Aug 20, 2017

The power rule only applies in its simplest form for powers of x.

Yes, d/dxcolor(red)x^n=ncolor(red)x^(n-1).

However, the chain rule changes this when there is an entire function being raised to a power.

In this case, d/dxcolor(blue)(f(x))^n=ncolor(blue)(f(x))^(n-1)*color(green)(f'(x)).

This is why we have d/dx(a-color(blue)((a^2-x^2))^(1/2))=-1/2color(blue)((a^2-x^2))^(-1/2)color(green)(d/dx(a^2-x^2).