Differentiate the function #f(x)= x/(x+1)# from first principle?

2 Answers
Apr 18, 2018

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

Explanation:

#color(blue)"differentiating from first principles"#

#f'(x)=lim_(hto0)(f(x+h)-f(x))/h#

#color(white)(f'(x))=lim_(hto0)((x+h)/(x+h+1)-x/(x+1))/h#

#"the aim now is to eliminate h from the denominator"#

#color(white)(f'(x))=lim_(hto0)((x+h)(x+1)-x(x+h+1))/(h(x+1)(x+h+1)#

#color(white)(f'(x))=lim_(hto0)(cancel(x^2)cancel(+hx)cancel(+x)+hcancel(-x^2)cancel(-hx)cancel(-x))/(h(x+1)(x+h+1))#

#color(white)(f'(x))=lim_(hto0)cancel(h)^1/(cancel(h)^1(x+1)(x+h+1)#

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

Apr 18, 2018

# d/dx x/(1+x) =1/(1+x)^2 #

Explanation:

The definition of the derivative of #y=f(x)# is

# f'(x)=lim_(h rarr 0) ( f(x+h)-f(x) ) / h #

So if # f(x) = x/(1+x) # then;

And so the derivative of #f(x)# is given by:

#f'(x) = lim_(h rarr 0) ( (x+h)/(1+(x+h)) - x/(1+x) ) /h #

# \ \ \ \ \ \ \ \ \ = lim_(h rarr 0) 1/h( (x+h)/(1+x+h) - x/(1+x) ) #

# \ \ \ \ \ \ \ \ \ = lim_(h rarr 0) 1/h( (x+h)(1+x)- x(1+x+h)) /((1+x+h)(1+x) ) #

# \ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( x+h + x^2+hx - x-x^2-hx) /(x(1+x+h)(1+x) ) #

# \ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( h ) /(h(1+x+h)(1+x) ) #

# \ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( 1 ) /((1+x+h)(1+x) ) #

# \ \ \ \ \ \ \ \ \ = ( 1 ) /((1+x+0)(1+x) ) #

# \ \ \ \ \ \ \ \ \ = ( 1 ) /((1+x)(1+x) ) #

# \ \ \ \ \ \ \ \ \ = ( 1 ) /(1+x)^2 #