1. ## implicit differentiation

The correlation between respiratory rate and body mass in the first three years of life can be expressed by the function
log R(w) = 1.78 - .38 log (w),
where w is the body weight (in kg) and R(w) is the respiratory rate (in breaths per minute). Find R'(w) using implicit differentiation.

2. Hello,
Originally Posted by Matho

The correlation between respiratory rate and body mass in the first three years of life can be expressed by the function
log R(w) = 1.78 - .38 log (w),
where w is the body weight (in kg) and R(w) is the respiratory rate (in breaths per minute). Find R'(w) using implicit differentiation.

Hum well... Differentiate both sides.

Use the chain rule for the left hand side !

$\displaystyle \frac{R'(w)}{R(w)}=-.38 \cdot \frac 1w$

So :
\displaystyle \begin{aligned} R'(w) &=\frac{-.38}{w} \cdot R(w) \\ &=\frac{-.38}{w} \cdot \exp \left(1.78-.38 \log(w)\right) \\ &=\frac{-.38}{w} \cdot \frac{e^{1.78}}{\exp \left(\log \left(w^{.38}\right)\right)} \quad (*) \\ &=\frac{-.38 e^{1.78}}{w \cdot w^{.38}} \\ &=\frac{-.38 e^{1.78}}{w^{1.38}} \end{aligned}

for (*) here is how I got it :
$\displaystyle \exp \left(1.78-.38 \log(w)\right)=\frac{\exp(1.78)}{\exp(.38 \log(w))}$
using the rule $\displaystyle e^{a-b}=\frac{e^a}{e^b}$

now using the rule $\displaystyle b \log(a)=\log \left(a^b\right)$, we get what I wrote.

As for what follows, just use the rules :
$\displaystyle \log(e^a)=a$
$\displaystyle e^a e^b=e^{a+b}$