Question #4ab09

1 Answer
Aug 15, 2017

#1/4#

Explanation:

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#