find equation for the normal line at prescribed point...

[Becky4paws]

Directions: The normal line to the curve y =f(x) atthe point P(x, f(x)) is the line perpendicular to the tangent line at P. Find an equation for the normal line to the given curve at the prescribed point.

y=(2/x) - square root of x; (1,1)

I'm just trying to find m, then I think I'll be ok.

y'=2x^-1 - x^-1/2
y'= -2x^-2 - (-1/2x^-3/2)
y'=-2/x^2 - (-1/2square root x ^3)

These square roots throw me. If that is correct I will need to substitute the value of x into the equation.
Then apply the formula y=m(x-x) +y

 

[pka]

You need to take another look at the derivative.
\(\displaystyle \L
y = \frac{2}{x} - \sqrt x \quad \Rightarrow \quad y' = - \frac{2}{{x^2 }} - \frac{1}{{2\sqrt x }}\)

 

[Becky4paws]

Next question

To solve and get m, would it be better to leave it as the equation or substitue the 1 in getting 2/1 - 1/2*squareroot 1?

 

[tkhunny]

How do you know the slope if you do not substitute the value x = 1 into the definition of the derivative?

 

[Becky4paws]

My attempt

y'=(-2/x^2) - (1/2squarerootx)

Substituting 1 for x:
y'(1) = (-2/1^2) - (1/2squareroot of 1)
y'(1) = -2 - 1/2
y'(1) = -5/2

to find the normal line:

y=-1/m(x-x)+y
y=(-5/2)(x-1) + 1
y= -5/2x + 5/2 + 2/2
y= -5/2x + 7/2

 

[pka]

Becky,
Look, if the tangent line has slope y’(1)=-5/2 then
the normal line has slope 2/5.