Question

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

Answer 1

The equation is `y = 1/6x +3/2`.

Explanation:

The derivative of `y= sqrt(x)` is `y' = 1/(2x^(1/2))`.

The slope of the tangent will be `1/(2(9)^(1/2)) = 1/6`

We can now find the equation of the line:

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

`y - 3 = 1/6(x - 9)`

`y - 3 = 1/6x - 3/2`

`y = 1/6x + 3/2`

Hopefully this helps!

Answer 2

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.

Explanation:

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

the slope of the line tangent to the graph of `f` at the point `(a,f(a))` is

`lim_(xrarra)(f(x)-f(a))/(x-a)` `" "` OR `" "` `lim_(hrarr0)(f(a+h)-f(a))/h`

First method

`lim_(xrarr9)(sqrtx-sqrt9)/(x-9) = lim_(xrarr9)(sqrtx-3)/(x-9)`

If we try to evaluate the limit by substitution, we get the indeterminate form `0/0`. We will rewrite the ratio and try again.

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

`(sqrtx-3)/(x-9)` by `1` in the form `(sqrtx+3)/(sqrtx+3)`

`(sqrtx-3)/(x-9) = ((sqrtx-3))/((x-9))((sqrtx+3))/((sqrtx+3))`

`= (x-9)/((x-9)(sqrtx+3))`

`= 1/(sqrtx+3)` `" "` (for `x != 9`).

The last equality is only for values of `x` other than `x=9`, but the limit doesn't care what happens when `x` is equal to `9`, it only cares what happens when `x` approaches `9`, so

`lim_(xrarr9)(sqrtx-3)/(x-9) = lim_(xrarr9)1/(sqrtx+3)`

`= 1/(sqrt9+3) = 1/(3+3)=1/6`

(Another way to rewrite `(sqrtx-3)/(x-9)` is to factor the denominator using the difference of squares factoring: `x-9 = (sqrtx+3)(sqrtx-3)`)

Second method

`lim_(hrarr0)(f(9+h)-f(9))/h = lim_(hrarr0) (sqrt(9+h)-3)/h`

Once again, substitution yields `0/0`. In this case it is hard to convince ourselves to factor the denimonator, so we'll rationalize the numerator and see what happens.

`lim_(hrarr0) (sqrt(9+h)-3)/h = lim_(hrarr0) ((sqrt(9+h)-3))/h * ((sqrt(9+h)+3))/((sqrt(9+h)+3))`

`= lim_(hrarr0)((9+h)-9)/(h(sqrt(9+h)+3))`

`= lim_(hrarr0)(h)/(h(sqrt(9+h)+3))`

`= lim_(hrarr0)1/(sqrt(9+h)+3)`

`= 1/(sqrt(9+0)+3) = 1/6`.

As I said, the other way discussed at the end of Method 1 is harder to use for this limit. But it is true that `h` can be "factored" as `h = (sqrt(9+h)+3)(sqrt(9+h)-3)` and then the ratio can be simplified.)

After you have found the slope

Proceed as in the answer by HSBC244.