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  1. #1
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    Question implicit differentiation

    I don't know how to solve this problem... Please help!

    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.

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  2. #2
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    Hello,
    Quote Originally Posted by Matho View Post
    I don't know how to solve this problem... Please help!

    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 !


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

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




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

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

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