How do you find the derivative of #f(x)= x+ sqrtx #?

Question

Christopher Agresta · Accepted Answer

#1+(1)/(2sqrt(x))#

We note that derivative of sums and differences can be split apart individually. Meaning,

#d/dx x# + #d/dx sqrt(x)#

where #d/dx x = 1# from the general power rule which states,

#d/dx x^n = nx^(n-1)#

Continuing on this basis,

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

Since we follow the same rules we first bring down the #1/2# and subtract #1# from #1/2# leaving,

#d/dx sqrt(x) = 1/2 x^(-1/2)#

We further note that we can remove the negative sign of a number by moving it to the opposite of its current location of the form #n/d# where we now move #x^(-1/2)# to the denominator resulting in #(1)/(2sqrt(x))#

Now we simply add the two derivatives together equaling,

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

Emma Aldridge · Answer

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

Note the sum rule for derivatives and the power rule:

#d/dx (f(x) + g(x)) = f'(x) + g'(x)#

As such, #d/dx f(x)#, treating #x# and #sqrt(x)# as their own functions in a sense, is...

#d/dx (x + sqrt(x))#

#d/dx x + d/dx sqrt(x)#

Using the power rule for the first term...

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

Using the power rule for the second...

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

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

Which can be rewritten as:

#1 + 1/(2sqrt(x))#