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

Question

Christopher Agresta · Accepted Answer

#ln3(3^t) - 3ln4(4^t)#

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 #(dg)/dt3^t# and then #(dg)/dt(4^t*3)#.

#(dg)/dt3^t# is easy - recall that #d/dxa^x = lna(a^x)#. That means #(dg)/dt3^t = ln3(3^t)#.

#(dg)/dt(4^t*3)# is a little more complicated, but bearable. Here's where we utilize the product rule. Remember, the product rule states that #d/dxuv = u'v+uv'#. In our case, #u = 4^t# and #v = 3#. Substituting,

#(dg)/dt(4^t*3) = (4^t)'(3)+(4^t)(3)'# #(dg)/dt(4^t*3) = 3ln4(4^t)+(4^t)(0)# #(dg)/dt(4^t*3) = 3ln4(4^t)#

Finally, we can now put it all together:

#(dg)/dt(3^t+4^t*3) = (dg)/dt3^t-(dg)/dt(4^t*3)# #(dg)/dt(3^t+4^t*3) = ln3(3^t) - 3ln4(4^t)#

Elijah Abram · Answer

undefined To use the product rule to find the derivative of $ g(t) = 3^t - 4^{t^3} $:

Let $ f(t) = 3^t $ and $ h(t) = 4^{t^3} $.

Now, apply the product rule:

$ g'(t) = f'(t) \cdot h(t) + f(t) \cdot h'(t) $.

Find the derivatives of $ f(t) $ and $ h(t) $:

$ f'(t) = (\ln(3) \cdot 3^t) $ and $ h'(t) = (\ln(4) \cdot 3t^2 \cdot 4^{t^3}) $.

Substitute these derivatives into the product rule:

$ g'(t) = (\ln(3) \cdot 3^t) \cdot (4^{t^3}) + (3^t) \cdot (\ln(4) \cdot 3t^2 \cdot 4^{t^3}) $.

So, the derivative of $ g(t) = 3^t - 4^{t^3} $ using the product rule is:

$ g'(t) = (\ln(3) \cdot 3^t) \cdot (4^{t^3}) + (3^t) \cdot (\ln(4) \cdot 3t^2 \cdot 4^{t^3}) $.