Let #f(x)# the square root of #x-2#. What is #f'(6)#?

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Answer 1 · 2018-01-24T15:45:22

# f'(6) = 1/4#

We have:

# f(x)=sqrt(x-2) #

So then differentiating wrt #x# we get:

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

So then:

# f'(6) = 1/(2sqrt(6-2))#
# \ \ \ \ \ \ \ \ = 1/(2sqrt(4))#
# \ \ \ \ \ \ \ \ = 1/4#

Answer 2 · 2018-01-24T15:51:42

#f'(6)=+-1/4#

Well, we got #f(x)=sqrt(x-2)#, so we first need to find the derivative of the function, that is #f'(x)#, before we can further solve the problem.

#f(x)=sqrt(x-2)#

Using the chain rule,

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

let #u=x-2#, then #du=dx#, #(du)/dx=1#, #f(u)=sqrt(u)#, and then #f'(u)=1/(2sqrt(u))# by the power rule.

So,

#f'(x)=f'(u)*(du)/(dx)#

#f'(x)=1/(2sqrt(u))*(du)/(dx)#

Substituting back, #u=x-2# and #(du)/(dx)=1#, we get

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

Since we have found #f'(x)#, we can plug in #x=6# to get #f'(6)#.

#:.f'(6)=1/(2sqrt(6-2))#

#f'(6)=1/(2sqrt(4)#

As #sqrt(4)=+-2#, we get two solutions for #f'(6)#.

#f'(6)=1/(+-2*2)#

#f'(6)=1/(+-4)#

#f'(6)=+-1/4#