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
aniket r.
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)#

mason m
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)#.

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