How to find the first derivative of this?

Question

$24000(1+\frac{0.061}{12})^{12t} + 11000\cdot e^{0.097t}$

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.

Answer 1

$$24000(1+\frac{0.061}{12})^{12t} + 11000\cdot e^{0.097t}$$ 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 $$(K^{12t})'=12 K^{12t} \ln K$$

Answer 2

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.