Question

What is the equation of the tangent line of `f(x) =sqrt(x+5)/(x^2-5x+5)` at `x=-1`?

Answer 1

the equation of the tangent is
`67x-484y+155=0`

Explanation:

Given:
`f(x)=(sqrt(x+5))/(x^2-5x+5)`
Let `y=f(x)`
We differentiate using the quotient rule in calculus
`y=(sqrt(x+5))/(x^2-5x+5)`
Let `u=sqrt(x+5)`

`v=x^2-5x+5`
`(du)/dx=1/(2sqrt(x+5)`
`(dv)/dx=2x-5`
`v^2=(x^2-5x+5)^2`

At `x=-1,`
`u=sqrt(-1+5)=sqrt4=2`
`v=(-1)^2-5(-1)+5=1+5+5=11`
`(du)/dx=1/(2sqrt(-1+5))=1/(2sqrt(4))=1/((2)(2))=1/4`
`(dv)/dx=2(-1)-5=-2-5=-7`

`y=u/v`
`y=2/11`
`(dy)/dx=d/dx(u/v)`
`d/dx(u/v)=(v(du)/dx-u(dv)/dx)/v^2`

Substituting
`d/dx(y)=((11)(1/4)-(2)(-7))/(11^2)`
`=(11/4+14)/121`
`=(11+56)/4(1/121)`
`(dy)/dx=67/484`

Slope of the tangent is `y'=67/484`
Thus,
At `x=-1, y=2/11, y'=67/484`

Equation of a straight line passing through the point `(a,b)`, and havihg slope `m` is given by
Here, `a=-1,b=2/11,m=67/484`
`(y-b)/(x-a)=m`

Substituting for x, y, m
`(y-2/11)/(x-(-1))=67/484`

Simplifying
`(11y-2)/(11(x+1))=67/484`
`(11y-2)/(x+1)=67/44`

Cross multiplying
`44(11y-2)=67(x+1)`
`484y-88=67x+67`

Rearranging
`67x-484y+67+88=0`

Expressing in the standard form
`67x-484y+155=0`
Thus, the equation of the tangent is
`67x-484y+155=0`