# How do you find the equation of a line tangent to #y=sqrtx# at (9,3)?

The equation is

We can now find the equation of the line:

Hopefully this helps!

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If you have learned to find derivatives using the power rule, then proceed as in the other answer. If you are using the definition of the slope of the tangent line, see below.

There are several possibilities for how you are learning to do this. I think two of the the more are:

First method

One method we can use is called "rationalizing the numerator". We'll multiply

Second method

After you have found the slope

Proceed as in the answer by HSBC244.

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To find the equation of a line tangent to y = √x at (9,3), we can use the derivative of the function. The derivative of y = √x is given by dy/dx = 1/(2√x).

To find the slope of the tangent line at (9,3), we substitute x = 9 into the derivative: dy/dx = 1/(2√9) = 1/6.

Since the slope of the tangent line is 1/6, we can use the point-slope form of a linear equation to find the equation of the line. The point-slope form is y - y₁ = m(x - x₁), where (x₁, y₁) is the given point and m is the slope.

Substituting the values, we have y - 3 = (1/6)(x - 9). Simplifying this equation gives the equation of the line tangent to y = √x at (9,3) as y = (1/6)x - 3/2.

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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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