1. ## Derivative

I'm having trouble finding the answer to this problem. : Find the derivative of the function 1/((the square root of(x)) and the equation of the line that is tangent to the graph of f and parallel to the given line 3x-y+1. I also found the answer to this one but it was kind of guess and check. : find the equations of two tangent lines to the graph of f that pass through the indicated point where f(x)=x^2 and the indicated point is (1, -3). The last question i have is how you guys put square roots and other symbols in. Can't seem to find them. Thanks a lot.

2. Originally Posted by jarny
I'm having trouble finding the answer to this problem. : Find the derivative of the function 1/((the square root of(x)) and the equation of the line that is tangent to the graph of f and parallel to the given line 3x-y+1. I also found the answer to this one but it was kind of guess and check. : find the equations of two tangent lines to the graph of f that pass through the indicated point where f(x)=x^2 and the indicated point is (1, -3). The last question i have is how you guys put square roots and other symbols in. Can't seem to find them. Thanks a lot.
The line $3x-y+1=0$ can be expressed as $y = 3x +1$. A parallel lines means the derivative has slopt $3$. Thus at which point is the derivative equal to 3?

3. Don't need an answer to the first one. I made a simple algebraic mistake which messed up a rather easy problem. Thanks for the answer to the second. Rather moronic of me not to see that Man that was stupid. That was a 1st grade level mistake. Just having a bad day thinking i guess. Can anyone tell me how to make the symbols though?

4. Originally Posted by jarny
find the equations of two tangent lines to the graph of f that pass through the indicated point where f(x)=x^2 and the indicated point is (1, -3).
$y = x^2$

$\Rightarrow y' =2x$

so the slope of any tangent line is given by $2x$

By the point-slope form, lines passing through $(1,-3)$ have the form:

$y + 3 = m(x - 1)$

$\Rightarrow y = mx - (m + 3)$

but we want lines where the slope is given by $2x$, so set $m = 2x$, we get:

$y = 2x^2 - (2x + 3) = 2x^2 - 2x - 3$

furthermore, we want lines that intersect with $y = x^2$, so equate the two functions. we get:

$x^2 = 2x^2 - 2x - 3$

$\Rightarrow x = 3$ or $x = -1$

let $m_1$ be the slope of the tangent line that intersects with $y = x^2$ at $x = 3$. thus $m_1 = 2x = 2(3) = 6$

so the first tangent line is: $\boxed {y = 6x - 9}$

in a similar fashion, we can find the tangent line that intersects with $y = x^2$ at $x = -1$ to be $\boxed {y = -2x - 1}$

The last question i have is how you guys put square roots and other symbols in. Can't seem to find them. Thanks a lot.
Use LaTex or if you don't want to, type sqrt(x) to mean $\sqrt {x}$