How do you find the derivative of #s=-3/(4t^2-2t+1)#?

Question

Ezra Adams · Accepted Answer

#(24t-6)/(4t^2-2t+1)^2# Through the quotient rule (see below).

The quotient rule states: #((Low*dHigh)-(High*dLow))/(Low)^2# So we need to determine the derivative of the top or the "dHigh" and the derivative of the bottom or the "dLow." The derivative of the top, or "dHigh" is the derivative of -3, which is 0. The derivative of the bottom, or "dLow" is the derivative of #4t^2−2t+1# From the Power Rule of Derivatives, we know the derivative of that is: #8t - 2# From this we can plug everything back into the above equation from the quotient rule. So we get the following after simplification: #(24t-6)/(4t^2-2t+1)^2#

Charles Anctil · Answer

undefined To find the derivative of $ s = -\frac{3}{4t^2 - 2t + 1} $, you would use the quotient rule, which states that if you have a function in the form $ \frac{f(x)}{g(x)} $, then its derivative is given by:

$$ \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} $$

Where $ f(x) = -3 $ and $ g(x) = 4t^2 - 2t + 1 $. Differentiate $ f(x) $ and $ g(x) $ with respect to $ t $, then apply the quotient rule. The derivative of $ f(x) $ with respect to $ t $ is 0 since it's a constant. The derivative of $ g(x) $ with respect to $ t $ is $ 8t - 2 $.

So applying the quotient rule:

$$ s' = \frac{0 \cdot (4t^2 - 2t + 1) - (-3) \cdot (8t - 2)}{(4t^2 - 2t + 1)^2} $$

$$ s' = \frac{6(4t - 1)}{(4t^2 - 2t + 1)^2} $$