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Answer 1 · 2018-01-31T06:02:35

#dy/dx=-1/x^2#

Given -

#f(x)=1/x#

#y=1/x#
The function can be written as -
#y=x^(-1)#
#dy/dx=(-1).x^(-1-1)#
#dy/dx=-x^(-2)#

#dy/dx=-1/x^2#

Answer 2 · 2018-01-31T07:08:57

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

According to definition of derivative for #f(x)#,

#f'(x)=(df)/(dx)=lim_(deltax->0)(f(x+deltax)-f(x))/(deltax)#

Here we have #f(x)=1/x#, therefore

#f(x+deltax)=1/(x+deltax)#

and #f'(x)=(df)/(dx)=lim_(deltax->0)(1/(x+deltax)-1/x)/(deltax)#

= #lim_(deltax->0)((x-x-deltax)/(x(x+deltax)))/(deltax)#

= #lim_(deltax->0)(-(deltax)/(x^2+xdeltax))/(deltax)#

= #lim_(deltax->0)-1/(x^2+xdeltax)#

= #-1/x^2#