Using the limit definition, how do you differentiate #f(x)=(x+1)^(1/2)#?

Question

1526 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-14T15:53:53

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

Based on the definition of derivative:

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

For #f(x) = (x+1)^(1/2) = sqrt(x+1)# we have then:

#d/dx ( sqrt(x+1) ) = lim_(h->0) ( sqrt(x+h+1)-sqrt(x+1))/h#

Now multiply numerator and denominator by the quantity:

#sqrt(x+h+1) + sqrt(x+1)#

and use the algebraic identity:

#(a-b)(a+b) = a^2-b^2#:

#d/dx ( sqrt(x+1) ) = lim_(h->0) (( sqrt(x+h+1)-sqrt(x+1))/h )((sqrt(x+h+1) + sqrt(x+1))/(sqrt(x+h+1) + sqrt(x+1)))#

#d/dx ( sqrt(x+1) ) = lim_(h->0) ( (x+h+1)-(x+1))/(h(sqrt(x+h+1) + sqrt(x+1))#

#d/dx ( sqrt(x+1) ) = lim_(h->0) cancel(h)/(cancel(h)(sqrt(x+h+1) + sqrt(x+1))#

#d/dx ( sqrt(x+1) ) = lim_(h->0) 1/(sqrt(x+h+1) + sqrt(x+1))#

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