How do you find the derivative of #f(x)=sqrt(a^2+x^2)#?

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Answer 1 · 2016-09-17T02:27:10

#f'(x) = x/(sqrt(a^2+x^2))#

The chain rule goes like this:
If #f(x) =(g(x))^n#, then #f'(x)=n(g(x))^(n-1)*d/dxg(x)#

Applying this rule:

#f(x) = sqrt(a^2+x^2)= (a^2+x^2)^(1/2)#

#f'(x) = 1/2(a^2+x^2)^(1/2-1) * d/dx(a^2+x^2)#

#f'(x) = 1/2(a^2+x^2)^(-1/2) * 2x#

#f'(x) = 1/(2(a^2+x^2)^(1/2))*2x#

#f'(x) = x/((a^2+x^2)^(1/2))#

#f'(x) = x/(sqrt(a^2+x^2))#