What is the slope of the tangent to #y=sqrt(1-x)# at #(-8, 3)#?

1 Answer
Apr 14, 2018

#m = -1/6#

Explanation:

Given #y=sqrt(1-x)#

Using the chain rule:

#dy/dx = dy/(dg)(dg)/dx#

where #y = sqrtg# and #g=1-x#, then #dy/(dg) = 1/(2sqrtg) = 1/(2sqrt(1-x))# and #(dg)/dx = -1#

Substituting into the chain rule:

#dy/dx = 1/(2sqrt(1-x)) -1#

#dy/dx = -1/(2sqrt(1-x))#

The slope, #m#, is the derivative evaluated at #x = -8#

#m = -1/(2sqrt(1--8))#

#m = -1/(2sqrt9)#

#m = -1/6#