Four questions in the Homework are bothering me :
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
$\displaystyle g(x) = \frac{27}{(2x+1)^3}$
$\displaystyle g'(x) = \frac{-3 \cdot 27}{(2x+1)^4}$
$\displaystyle g'(1) = \frac{-3 \cdot 27}{(2 \cdot 1 +1)^4} =$ $\displaystyle -\frac{81}{81} = -1$
-Dan
Let $\displaystyle f(x) = \frac{6}{5 \sqrt{3x^4} - x}$. Find f'(1).
$\displaystyle f(x) = \frac{6}{5 \sqrt{3x^4} - x}$
$\displaystyle f'(x) = \frac{-6}{(5 \sqrt{3x^4} - x)^{2}} \cdot \left ( 5 \cdot \frac{1}{2\sqrt{3x^4}} \cdot 3 \cdot 4x^3 - 1 \right )$
$\displaystyle f'(x) = \frac{-6 \cdot (10\sqrt{3}x - 1)}{(5\sqrt{3}x^2 - x)^2}$
$\displaystyle f'(1) = \frac{-6 \cdot (10\sqrt{3} \cdot 1 - 1)}{(5\sqrt{3}(1)^2 - 1)^2} \approx -1.66878$
-Dan
$\displaystyle s = 5 - 3t - \left [ t^{-2} + (2t - 5)^5 \right ] ^4$
$\displaystyle s' = -3 - 4 \left [ t^{-2} + (2t - 5)^5 \right ] ^3 \cdot \left ( -2t^{-3} + 5(2t - 5)^4 \cdot 2 \right )$
$\displaystyle s' = -3 - 4 \left ( -2t^{-3} + 10(2t - 5)^4 \right ) \left [ t^{-2} + (2t - 5)^5 \right ] ^3 $
-Dan
$\displaystyle V(t) = 50000 \left ( 1 - \frac{t}{30} \right ) ^2$
$\displaystyle V'(t) = 50000 \cdot 2 \left ( 1 - \frac{t}{30} \right ) \cdot \frac{-1}{30}$
$\displaystyle V'(t) = - \frac{10000}{3} \left ( 1 - \frac{t}{30} \right )$
$\displaystyle V'(10) = - \frac{10000}{3} \left ( 1 - \frac{10}{30} \right )$
$\displaystyle V'(10) = - \frac{20000}{9}$
Technically they want 20000/9 gal/min (ie not negative) because the question asked for the rate the water is flowing out.
-Dan