How to find the first derivative of this? $24000(1+\frac{0.061}{12})^{12t}  + 11000\cdot e^{0.097t}$
I'm trying to find the derivative of this but keep getting confused on the steps. Can anyone explain the steps and solution to this problem. 
 A: $$24000(1+\frac{0.061}{12})^{12t}  + 11000\cdot e^{0.097t}$$
For simplicity I substitute..
$$K=(1+\frac{0.061}{12})$$
$$24000K^{12t}  + 11000\cdot e^{0.097t}$$
Note that
$$(a^{bt})'=((e^{\ln a})^{bt})'=(e^{bt \ln a})'=e^{bt \ln a} b\ln a=a^{bt} b\ln a$$
So
$$(K^{12t})'=12 K^{12t}  \ln K$$
A: I'm assuming that, by "24000(1+0.061/12)^12t  + 11000e^0.097t" you mean
$$24000\left(1+\frac{0.061}{12}\right)^{12t}+11000e^{0.097t}$$
Instead of giving you the full-on answer, I'm going to walk you through a step-by-step derivation of a general formula.
Suppose that $f(t)=a_1(a_2)^{a_3t}+a_4e^{a_5t}$, where $a_1,a_2,a_3,a_4,a_5$ are all constants. Thus:
$$f'(t)=\frac{d}{dt}\Biggl(a_1(a_2)^{a_3t}+a_4e^{a_5t}\Biggr)$$
Therefore: $$f'(t)=\frac{d}{dt}a_1(a_2)^{a_3t}+\frac{d}{dt}a_4e^{a_5t}$$
$$f'(t)=a_1\frac{d}{dt}(a_2)^{a_3t}+a_4\frac{d}{dt}e^{a_5t}$$
Now, we define $p=\frac{d}{dt}(a_2)^{a_3t}$, and $q=\frac{d}{dt}e^{a_5t}$. Thus:
$$f'(t)=a_1p+a_4q$$
Next, lets find $$p=\frac{d}{dt}(a_2)^{a_3t}$$ 
Substituting in $u=a_3t$,$$p=\frac{d}{dt}a_2^u$$
And using the chain rule along with $\frac{d}{dt}a^t=a^t \ln a$, we arrive at
$$p=\ln(a_2)a_2^u\cdot \frac{d}{dt}u$$
$$p=\ln(a_2)a_2^{a_3}\frac{d}{dt}a_3t$$
$$p=a_3\ln(a_2)a_2^{a_3t}$$
Thus, $$f'(t)=a_1a_3\ln(a_2)a_2^{a_3t}+a_4q$$
Okay next step:
$$q=\frac{d}{dt}e^{a_5t}$$
Again, using the chain rule along with $\frac{d}{dt}a^t=a^t \ln a$, but this time noting that $\ln(e)=1$,
$$q=a_5e^{a_5t}$$
When we plug back in,
$$f'(t)=a_1a_3\ln(a_2)a_2^{a_3t}+a_4a_5e^{a_5t}$$
Which is the formula. If you plug the values $a_1=24000$, $a_2=1+\frac{0.061}{12}$, $a_3=12$, $a_4=11000$, and $a_5=0.097$ into the equation for $f'(t)$ that we just found, you'll arrive at your answer.
