What is the equation of the tangent line of #f(x) =sqrt(x+15)-x/(x-15)# at #x=5#?

Question

1534 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-15T12:22:31

#20y-(3+sqrt5)x+5-35sqrt5=0#

#y=sqrt(x+15)-x/(x-15)#

Substitute for #x# value to get the coordinates of the point which lies on the tangent

#y=(1+4sqrt5)/2#

Differentiate the function Using

#y'=1/(2sqrt(x+15))-((x-15)xx1-x xx1)/(x-15)^2#

Simplify

#y'=1/(2sqrt(x+15))+15/(x-15)^2#

Now We substitute for x=5 to get the Slope of the Tangent #m#

#y'_(x=5)=(3+sqrt5)/20#=#m#

now Substitute in the straight line equation

#y-y_1=m(x-x_1)#

#y-(1+4sqrt5)/2=(3+sqrt5)/20(x-5)#

#20y-(3+sqrt5)x+5-35sqrt5=0#