# Chain Rule

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• Dec 6th 2006, 02:06 PM
aaasssaaa
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Chain Rule
Four questions in the Homework are bothering me:mad: :

see attachment cause I had to use word to put the equation
• Dec 6th 2006, 02:48 PM
galactus
For the most part, it's just differentiation. Find the derivative and enter in the given values.
• Dec 6th 2006, 03:34 PM
aaasssaaa
Yea this is what I tried to do except i kept getting the wrong answer
• Dec 6th 2006, 04:12 PM
topsquark
Quote:

Originally Posted by aaasssaaa
Four questions in the Homework are bothering me:mad: :

see attachment cause I had to use word to put the equation

let g(x) = 27/(2x+1)^3 Determine the slope of the tangent to the curve at x=1

$g(x) = \frac{27}{(2x+1)^3}$

$g'(x) = \frac{-3 \cdot 27}{(2x+1)^4}$

$g'(1) = \frac{-3 \cdot 27}{(2 \cdot 1 +1)^4} =$ $-\frac{81}{81} = -1$

-Dan
• Dec 6th 2006, 04:21 PM
topsquark
Quote:

Originally Posted by aaasssaaa
Four questions in the Homework are bothering me:mad: :

see attachment cause I had to use word to put the equation

Let $f(x) = \frac{6}{5 \sqrt{3x^4} - x}$. Find f'(1).

$f(x) = \frac{6}{5 \sqrt{3x^4} - x}$

$f'(x) = \frac{-6}{(5 \sqrt{3x^4} - x)^{2}} \cdot \left ( 5 \cdot \frac{1}{2\sqrt{3x^4}} \cdot 3 \cdot 4x^3 - 1 \right )$

$f'(x) = \frac{-6 \cdot (10\sqrt{3}x - 1)}{(5\sqrt{3}x^2 - x)^2}$

$f'(1) = \frac{-6 \cdot (10\sqrt{3} \cdot 1 - 1)}{(5\sqrt{3}(1)^2 - 1)^2} \approx -1.66878$

-Dan
• Dec 6th 2006, 04:26 PM
topsquark
Quote:

Originally Posted by aaasssaaa
Four questions in the Homework are bothering me:mad: :

see attachment cause I had to use word to put the equation

$s = 5 - 3t - \left [ t^{-2} + (2t - 5)^5 \right ] ^4$

$s' = -3 - 4 \left [ t^{-2} + (2t - 5)^5 \right ] ^3 \cdot \left ( -2t^{-3} + 5(2t - 5)^4 \cdot 2 \right )$

$s' = -3 - 4 \left ( -2t^{-3} + 10(2t - 5)^4 \right ) \left [ t^{-2} + (2t - 5)^5 \right ] ^3$

-Dan
• Dec 6th 2006, 04:31 PM
topsquark
Quote:

Originally Posted by aaasssaaa
Four questions in the Homework are bothering me:mad: :

see attachment cause I had to use word to put the equation

$V(t) = 50000 \left ( 1 - \frac{t}{30} \right ) ^2$

$V'(t) = 50000 \cdot 2 \left ( 1 - \frac{t}{30} \right ) \cdot \frac{-1}{30}$

$V'(t) = - \frac{10000}{3} \left ( 1 - \frac{t}{30} \right )$

$V'(10) = - \frac{10000}{3} \left ( 1 - \frac{10}{30} \right )$

$V'(10) = - \frac{20000}{9}$

Technically they want 20000/9 gal/min (ie not negative) because the question asked for the rate the water is flowing out.

-Dan
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