The slope of a function's tangent line at a point is given by the function's derivative. So, we need to find the derivative of #sqrt(x-5)=(x-5)^(1/2)#.
Note that #d/dxx^(1/2)=1/2x^(-1/2)#. So, the chain rule states that if we have a function to the #1//2# power instead, we follow the same process of the power rule, by moving down the #1//2# as a multiplicative constant and reducing the power by #1# #(#here, to get to #-1//2)#, but we also multiply by the derivative of the inner function.
Thus, #d/dx(f(x))^(1/2)=1/2(f(x))^(-1/2)*f'(x)#.
So, the derivative of #(x-5)^(1/2)# is:
#d/dx(x-5)^(1/2)=1/2(x-5)^(-1/2)*d/dx(x-5)#
The derivative of #x-5# is just #1#, so:
#dy/dx=1/2(x-5)^(-1/2)=1/(2sqrt(x-5))#
The slope of the tangent line to the original function #y# is found by plugging #x=9# into #dy//dx#:
#1/(2sqrt(9-5))=color(red)(1/4#
