Chain rule

**Mathmania** · December 8th 2006, 12:13 PM

Atachement

**topsquark** · December 8th 2006, 12:38 PM

Originally Posted by Mathmania

Atachement

$y = \frac{\sqrt{3x^2+4}}{(x^2-3)^3}$

Recall that if we have y = f(g(x)) that
$y' = f'(g(x)) \cdot g'(x)$

And if y = f(g(h(x)))
$y' = f'(g(h(x)) \cdot g'(h(x)) \cdot h'(x)$

etc, etc.

So to facilitate this process a bit I'm going to rewrite this as:
$y = \sqrt{3x^2+4}(x^2-3)^{-3}$

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.

$y' = \left ( \frac{\frac{1}{2}}{\sqrt{3x^2+4)}} \cdot (6x) \right ) (x^2 - 3)^{-3} +$ $\sqrt{3x^2+4} \cdot \left ( (-3) \cdot (x^2-3)^{-4} \cdot (2x) \right )$

After a bit of simplifying I get:
$y' = \frac{-3x(5x^2+11)\sqrt{3x^2+4}}{(x^2-3)^4(3x^2+4)}$

Now. At the point (2, 4) the slope has a value of -93/2. Thus the tangent line will be:
$y = -\frac{93}{2}x + b$
and will pass through the point (2, 4):

$4 = -\frac{93}{2} \cdot 2 + b$

$4 = -93 + b$

$b = 97$

Thus the tangent line is
$y = -\frac{93}{2}x + 97$

-Dan

**topsquark** · December 8th 2006, 12:48 PM

Originally Posted by Mathmania

Atachement

3. $g(x) = \sqrt{(4x+5)^3 - 1}$

$g'(x) = \frac{ \frac{1}{2} }{\sqrt{(4x+5)^3 - 1}} \cdot 3(4x+5)^2 \cdot 4$

$g'(x) = \frac{6(4x+5)^2}{\sqrt{(4x+5)^3 - 1}}$

-Dan

**topsquark** · December 8th 2006, 01:07 PM

Originally Posted by Mathmania

Atachement

Like the first problem we need to have the value of the slope of the function. So:
$y = \frac{(x^2 + 1)^3}{(x^2 - 1)^3}$

I'll use the quotient rule here, then the chain rule for each factor:
$y' = \frac{ (3(x^2+1)^2 \cdot (2x))(x^2 - 1)^3 - (x^2+1)^3(3(x^2 - 1)^2 \cdot (2x)) }{\left ( (x - 1)^3 \right ) ^2}$

After a bit of simplification:
$y' = - \frac{12x(x + 1)^2}{(x - 1)^4}$

Now for the first tangent line. We want the tangent at the point $\left ( -3, \frac{1000}{512} \right )$ . The slope will be:
$y' = - \frac{12(-3)((-3)^2 + 1)^2}{((-3)^2 - 1)^4} = \frac{225}{256}$

So the first line has the form $y = \frac{225}{256}x + b$ . Inserting the point $\left ( -3, \frac{1000}{512} \right )$ into this I get that $b = \frac{1175}{256}$ .

So the first line is: $y = \frac{225}{256}x + \frac{1175}{256}$ .

In a similar fashion the line tangent to the function at the point $\left ( -2, \frac{125}{27} \right )$ is: $y = \frac{200}{27}x + \frac{175}{9}$ .

To find the point of intersection we need to solve the system of equations:
$y = \frac{225}{256}x + \frac{1175}{256}$

$y = \frac{200}{27}x + \frac{175}{9}$

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: $\left ( - \frac{4107}{1805}, \frac{935}{361} \right )$ .

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

**Soroban** · December 8th 2006, 01:31 PM

Hello, Mathmania!

1) Find the equation of the tangent line to the curve

. . $y \:=\:\frac{\sqrt{3x^2+4}}{(x^2-3)^3}$ at the point $(2,4)$ .

I would use Log Differentiation . . .

We have: . $\ln y \;=\;\frac{1}{2}\ln(3x^2+4) - 3\ln(x^2-3)$

. . Then: . $\frac{1}{y}\!\cdot\!y' \;=\;\frac{1}{2}\!\cdot\!\frac{6x}{3x^2+4} - 3\!\cdot\!\frac{2x}{x^3-3} \quad\Rightarrow\quad y' \;=\;y\left[\frac{3x}{3x^2+4} - \frac{6x}{x^2-3}\right]$

At $(2,4):\;y' \;=\;4\left[\frac{6}{16} - \frac{12}{1}\right] \;=\;4\left(-\frac{93}{8}\right)\;=\;-\frac{93}{2}$

We have: . $y - 4 \:=\:-\frac{93}{2}(x - 2)\quad\Rightarrow\quad\boxed{y \;=\;-\frac{93}{2}x + 97}$

Given: $f(x)\:=\:4x+5$ and $g(x) \:=\:\sqrt{\left[f(x)\right]^3 - 1}$
. . Determine: . $g'(x).$

We have: . $g(x) \;=\;\left[(4x+5)^3 - 1\right]^{\frac{1}{2}}$

Chain Rule: . $g'(x) \;=\;\frac{1}{2}\left[(4x+5)^3-1\right]^{-\frac{1}{2}}\cdot3(4x+5)^2\cdot4$

Therefore: . $\boxed{g'(x)\;=\;\frac{6(4x+5)^2}{\sqrt{(4x+5)^3-1}}}$

**Mathmania** · December 8th 2006, 01:40 PM

for the last part in question 2 can u show me what method you used, i tried using 2 different methods but i get a weird answer

**topsquark** · December 8th 2006, 01:51 PM

Originally Posted by Mathmania

for the last part can u show me what method you used, i tried using 2 different methods but i get a weird answer

nevermind got it

I'm presuming this message is directed at me. I'm going to post the solution anyway, just in case someone wants it.

$y = \frac{225}{256}x + \frac{1175}{256}$

$y = \frac{200}{27}x + \frac{175}{9}$

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:
$\frac{225}{256}x + \frac{1175}{256} = \frac{200}{27}x + \frac{175}{9}$

$\left ( \frac{225}{256} - \frac{200}{27} \right ) x = \frac{175}{9} - \frac{1175}{256}$

$- \frac{45125}{6912} x = \frac{34225}{2304}$

$x = - \frac{34225}{2304} \cdot \frac{6912}{45125} = - \frac{4107}{1805}$

Then use either of the original equations to find y. I'll use the first:
$y = - \frac{225}{256} \cdot \frac{4107}{1805} + \frac{1175}{256}$

$y = - \frac{184815}{92416} + \frac{1175}{256} = \frac{935}{361}$

-Dan

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