How do you find the derivative of #y=1/(2x^2+1)#?

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Answer 1 · 2018-02-06T00:35:42

#f'(x)=-(4x)/(2x^2+1)^2#

First, let's rewrite this function.
#1/(2x^2+1)=>(2x^2+1x^0)^-1#

Now, remember the power rule and the chain rule.

Power rule: #d/dx(x^n)=nx^(n-1)# where #n# is a constant.

Chain rule: If #f(x)=g(h(x))#, then #f'(x)=g'(h(x))*h'(x)#

Therefore, we have:

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

=>#-(2x^2+1)^-2*4x^(2-1)+0x^-1#

=>#-(2x^2+1)^-2*4x^(1)+0#

=>#-1/(2x^2+1)^2*4x#

=>#-(4x)/(2x^2+1)^2#

That is our answer!