# Simple differentiation question

#### downthesun01

$$\displaystyle s(t)=\frac{a(t)'}{a(t)}$$

If

$$\displaystyle s(t)=\frac{4}{t+3}$$

Find a(t)

To me a(t) would just be $$\displaystyle (t+3)^4$$

But the book has $$\displaystyle \frac{t+3}{3}^4$$

How'd the book get that answer?

#### Prove It

MHF Helper
$$\displaystyle s(t)=\frac{a(t)'}{a(t)}$$

If

$$\displaystyle s(t)=\frac{4}{t+3}$$

Find a(t)

To me a(t) would just be $$\displaystyle (t+3)^4$$

But the book has $$\displaystyle \frac{t+3}{3}^4$$

How'd the book get that answer?
\displaystyle \displaystyle \begin{align*} \frac{4}{t + 3} &= \frac{1}{a}\, \frac{da}{dt} \\ \int{\frac{4}{t + 3}\,dt} &= \int{\frac{1}{a}\,\frac{da}{dt}\,dt} \\ 4\ln{|t + 3|} + C_1 &= \int{\frac{1}{a}\,da} \\ 4\ln{|t + 3|} + C_1 &= \ln{|a|} + C_2 \\ 4\ln{|t + 3|} + C_1 - C_2 &= \ln{|a|} \\ e^{4\ln{|t+3|} + C_1 - C_2} &= |a| \\ e^{C_1 - C_2} e^{\ln{\left|t +3 \right|^4}} &= |a| \\ a &= C \left( t + 3 \right) ^4 \textrm{ where }C = \pm e^{C_1 - C_2} \end{align*}

• downthesun01 and (deleted member)