Atachement
$\displaystyle y = \frac{\sqrt{3x^2+4}}{(x^2-3)^3}$
Recall that if we have y = f(g(x)) that
$\displaystyle y' = f'(g(x)) \cdot g'(x)$
And if y = f(g(h(x)))
$\displaystyle y' = f'(g(h(x)) \cdot g'(h(x)) \cdot h'(x)$
etc, etc.
So to facilitate this process a bit I'm going to rewrite this as:
$\displaystyle y = \sqrt{3x^2+4}(x^2-3)^{-3}$
We are going to be using the product rule: y = f(x)g(x) --> y' = f'(x)g(x) + f(x)g'(x). Then we use the chain rule on the individual factors.
$\displaystyle y' = \left ( \frac{\frac{1}{2}}{\sqrt{3x^2+4)}} \cdot (6x) \right ) (x^2 - 3)^{-3} + $ $\displaystyle \sqrt{3x^2+4} \cdot \left ( (-3) \cdot (x^2-3)^{-4} \cdot (2x) \right )$
After a bit of simplifying I get:
$\displaystyle y' = \frac{-3x(5x^2+11)\sqrt{3x^2+4}}{(x^2-3)^4(3x^2+4)}$
Now. At the point (2, 4) the slope has a value of -93/2. Thus the tangent line will be:
$\displaystyle y = -\frac{93}{2}x + b$
and will pass through the point (2, 4):
$\displaystyle 4 = -\frac{93}{2} \cdot 2 + b$
$\displaystyle 4 = -93 + b$
$\displaystyle b = 97$
Thus the tangent line is
$\displaystyle y = -\frac{93}{2}x + 97$
-Dan
Like the first problem we need to have the value of the slope of the function. So:
$\displaystyle y = \frac{(x^2 + 1)^3}{(x^2 - 1)^3}$
I'll use the quotient rule here, then the chain rule for each factor:
$\displaystyle y' = \frac{ (3(x^2+1)^2 \cdot (2x))(x^2 - 1)^3 - (x^2+1)^3(3(x^2 - 1)^2 \cdot (2x)) }{\left ( (x - 1)^3 \right ) ^2}$
After a bit of simplification:
$\displaystyle y' = - \frac{12x(x + 1)^2}{(x - 1)^4}$
Now for the first tangent line. We want the tangent at the point $\displaystyle \left ( -3, \frac{1000}{512} \right )$. The slope will be:
$\displaystyle y' = - \frac{12(-3)((-3)^2 + 1)^2}{((-3)^2 - 1)^4} = \frac{225}{256}$
So the first line has the form $\displaystyle y = \frac{225}{256}x + b$. Inserting the point $\displaystyle \left ( -3, \frac{1000}{512} \right )$ into this I get that $\displaystyle b = \frac{1175}{256}$.
So the first line is: $\displaystyle y = \frac{225}{256}x + \frac{1175}{256}$.
In a similar fashion the line tangent to the function at the point $\displaystyle \left ( -2, \frac{125}{27} \right )$ is: $\displaystyle y = \frac{200}{27}x + \frac{175}{9}$.
To find the point of intersection we need to solve the system of equations:
$\displaystyle y = \frac{225}{256}x + \frac{1175}{256}$
$\displaystyle y = \frac{200}{27}x + \frac{175}{9}$
Use your favorite method here. It's ugly and if you want help I'll provide it. Until then I'll simply state the solution:
The point of intersection is: $\displaystyle \left ( - \frac{4107}{1805}, \frac{935}{361} \right )$.
Two comments:
1) Check my math. This is sufficiently complicated I may have made a small error somewhere. I'd simply graph the problem, but my nicer graphing software isn't available. (On my TI-92 it seems to check out, but the resolution is bad.)
2) Whoever came up with this one is a sadist!
-Dan
Hello, Mathmania!
1) Find the equation of the tangent line to the curve
. . $\displaystyle y \:=\:\frac{\sqrt{3x^2+4}}{(x^2-3)^3}$ at the point $\displaystyle (2,4)$.
I would use Log Differentiation . . .
We have: .$\displaystyle \ln y \;=\;\frac{1}{2}\ln(3x^2+4) - 3\ln(x^2-3) $
. . Then: .$\displaystyle \frac{1}{y}\!\cdot\!y' \;=\;\frac{1}{2}\!\cdot\!\frac{6x}{3x^2+4} - 3\!\cdot\!\frac{2x}{x^3-3} \quad\Rightarrow\quad y' \;=\;y\left[\frac{3x}{3x^2+4} - \frac{6x}{x^2-3}\right] $
At $\displaystyle (2,4):\;y' \;=\;4\left[\frac{6}{16} - \frac{12}{1}\right] \;=\;4\left(-\frac{93}{8}\right)\;=\;-\frac{93}{2} $
We have: . $\displaystyle y - 4 \:=\:-\frac{93}{2}(x - 2)\quad\Rightarrow\quad\boxed{y \;=\;-\frac{93}{2}x + 97}$
Given: $\displaystyle f(x)\:=\:4x+5$ and $\displaystyle g(x) \:=\:\sqrt{\left[f(x)\right]^3 - 1}$
. . Determine: .$\displaystyle g'(x).$
We have: .$\displaystyle g(x) \;=\;\left[(4x+5)^3 - 1\right]^{\frac{1}{2}} $
Chain Rule: .$\displaystyle g'(x) \;=\;\frac{1}{2}\left[(4x+5)^3-1\right]^{-\frac{1}{2}}\cdot3(4x+5)^2\cdot4$
Therefore: .$\displaystyle \boxed{g'(x)\;=\;\frac{6(4x+5)^2}{\sqrt{(4x+5)^3-1}}}$
I'm presuming this message is directed at me. I'm going to post the solution anyway, just in case someone wants it.
$\displaystyle y = \frac{225}{256}x + \frac{1175}{256}$
$\displaystyle y = \frac{200}{27}x + \frac{175}{9}$
Note: A graphing calculator would be very handy here. If you don't have one I would recommend using decimals instead of trying to combine the fractions by hand. (This is one of the rare places I suggest that.)
We are looking for a place where x and y are the same for both equations. Since both are solved for y, we can say that the right hand sides of both equations are equal:
$\displaystyle \frac{225}{256}x + \frac{1175}{256} = \frac{200}{27}x + \frac{175}{9}$
$\displaystyle \left ( \frac{225}{256} - \frac{200}{27} \right ) x = \frac{175}{9} - \frac{1175}{256}$
$\displaystyle - \frac{45125}{6912} x = \frac{34225}{2304}$
$\displaystyle x = - \frac{34225}{2304} \cdot \frac{6912}{45125} = - \frac{4107}{1805}$
Then use either of the original equations to find y. I'll use the first:
$\displaystyle y = - \frac{225}{256} \cdot \frac{4107}{1805} + \frac{1175}{256}$
$\displaystyle y = - \frac{184815}{92416} + \frac{1175}{256} = \frac{935}{361}$
-Dan