How do you use the product Rule to find the derivative of g(t) = (3^t) - (4^t3)?

Sep 22, 2015

$\ln 3 \left({3}^{t}\right) - 3 \ln 4 \left({4}^{t}\right)$

Explanation:

We'll have to use a little more than the product rule to find this derivative. Luckily, we can take it one step at a time - finding $\frac{\mathrm{dg}}{\mathrm{dt}} {3}^{t}$ and then $\frac{\mathrm{dg}}{\mathrm{dt}} \left({4}^{t} \cdot 3\right)$.

$\frac{\mathrm{dg}}{\mathrm{dt}} {3}^{t}$ is easy - recall that $\frac{d}{\mathrm{dx}} {a}^{x} = \ln a \left({a}^{x}\right)$. That means $\frac{\mathrm{dg}}{\mathrm{dt}} {3}^{t} = \ln 3 \left({3}^{t}\right)$.

$\frac{\mathrm{dg}}{\mathrm{dt}} \left({4}^{t} \cdot 3\right)$ is a little more complicated, but bearable. Here's where we utilize the product rule. Remember, the product rule states that $\frac{d}{\mathrm{dx}} u v = u ' v + u v '$. In our case, $u = {4}^{t}$ and $v = 3$. Substituting,

$\frac{\mathrm{dg}}{\mathrm{dt}} \left({4}^{t} \cdot 3\right) = \left({4}^{t}\right) ' \left(3\right) + \left({4}^{t}\right) \left(3\right) '$
$\frac{\mathrm{dg}}{\mathrm{dt}} \left({4}^{t} \cdot 3\right) = 3 \ln 4 \left({4}^{t}\right) + \left({4}^{t}\right) \left(0\right)$
$\frac{\mathrm{dg}}{\mathrm{dt}} \left({4}^{t} \cdot 3\right) = 3 \ln 4 \left({4}^{t}\right)$

Finally, we can now put it all together:

$\frac{\mathrm{dg}}{\mathrm{dt}} \left({3}^{t} + {4}^{t} \cdot 3\right) = \frac{\mathrm{dg}}{\mathrm{dt}} {3}^{t} - \frac{\mathrm{dg}}{\mathrm{dt}} \left({4}^{t} \cdot 3\right)$
$\frac{\mathrm{dg}}{\mathrm{dt}} \left({3}^{t} + {4}^{t} \cdot 3\right) = \ln 3 \left({3}^{t}\right) - 3 \ln 4 \left({4}^{t}\right)$