# How do you find the equation of a line normal to the function #y=5/sqrtx-sqrtx# at (1,4)?

The equation is

This has derivative

Hence, the slope of the tangent is:

We now have the equation as:

Hopefully this helps!

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To find the equation of a line normal to the function y = 5/√x - √x at (1,4), we need to determine the slope of the normal line.

First, let's find the derivative of the given function.

Differentiating y = 5/√x - √x with respect to x, we get:

dy/dx = (d/dx)(5/√x) - (d/dx)(√x) = (-5/2)x^(-3/2) - (1/2)x^(-1/2) = -5/(2√x^3) - 1/(2√x)

Now, let's substitute x = 1 into the derivative to find the slope at (1,4):

dy/dx |(x=1) = -5/(2√1^3) - 1/(2√1) = -5/2 - 1/2 = -6/2 = -3

The slope of the normal line is the negative reciprocal of the derivative at (1,4). Therefore, the slope of the normal line is 1/3.

Using the point-slope form of a line, we can write the equation of the line as:

y - y1 = m(x - x1)

Substituting the values of (x1, y1) = (1, 4) and m = 1/3, we have:

y - 4 = (1/3)(x - 1)

Simplifying the equation, we get:

3y - 12 = x - 1

Rearranging the equation, we find the equation of the line normal to the function y = 5/√x - √x at (1,4) as:

x - 3y + 11 = 0

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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