How do you find the derivative of # 1/(x^2-1)# using the limit definition?

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Answer 1 · 2016-09-19T06:49:29

#(df)/(dx)=(-2x)/(x^2-1)^2#

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

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

Hence, #f(x+h)-f(x)=1/((x+h)^2-1)-1/(x^2-1)#

= #((x^2-1)-((x+h)^2-1))/((x^2-1)((x+h)^2-1))#

= #((x^2-1)-(x^2+2hx+h^2-1))/((x^2-1)((x+h)^2-1))#

= #((x^2-1-x^2-2hx-h^2+1))/((x^2-1)((x+h)^2-1))#

= #((-2hx-h^2))/((x^2-1)((x+h)^2-1))# and

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

Now #(df)/(dx)=Lt_(h->0)(f(x+h)-f(x))/h#

= #Lt_(h->0)((-2x-h))/((x^2-1)((x+h)^2-1))#

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

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