Derivatives
- Thread starter Dazed
- Start date
G'day, Dazed.
We have
. . \(\displaystyle \L y = \frac{114.44}{1 \, + \, 0.0383e^{1.05x}}\)
The quotient rule would give
. . \(\displaystyle \L \begin{align*}
\\
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\\
y' &= \frac{(0)(1 + 0.0383e^{1.05x}) - (0.040215e^{1.05x})(114.44)}{(1 + 0.0383e^{1.05x})^2} \\
\\
\\
\\
\\
&= \frac{- 114.44(0.040215e^{1.05x})}{(1 + 0.0383e^{1.05x})^2} \\
\end{align*}\)
Alternatively we could apply the chain rule to
. . . \(\displaystyle \L
\\
\\
\\
y = 114.44(1 \, + \, 0.0383e^{1.05x})^{-1}\)
To get
. . \(\displaystyle \L \begin{align*}
\\
\\
\\
y' &= -114.44(1 \, + \, 0.0383e^{1.05x})^{-2} \cdot (0.040215e^{1.05x}) \\
\\
\\
\\
\\
&= \frac{- 114.44(0.040215e^{1.05x})}{(1 + 0.0383e^{1.05x})^2} \\
\end{align*}\)
We have
. . \(\displaystyle \L y = \frac{114.44}{1 \, + \, 0.0383e^{1.05x}}\)
The quotient rule would give
. . \(\displaystyle \L \begin{align*}
\\
\\
\\
y' &= \frac{(0)(1 + 0.0383e^{1.05x}) - (0.040215e^{1.05x})(114.44)}{(1 + 0.0383e^{1.05x})^2} \\
\\
\\
\\
\\
&= \frac{- 114.44(0.040215e^{1.05x})}{(1 + 0.0383e^{1.05x})^2} \\
\end{align*}\)
Alternatively we could apply the chain rule to
. . . \(\displaystyle \L
\\
\\
\\
y = 114.44(1 \, + \, 0.0383e^{1.05x})^{-1}\)
To get
. . \(\displaystyle \L \begin{align*}
\\
\\
\\
y' &= -114.44(1 \, + \, 0.0383e^{1.05x})^{-2} \cdot (0.040215e^{1.05x}) \\
\\
\\
\\
\\
&= \frac{- 114.44(0.040215e^{1.05x})}{(1 + 0.0383e^{1.05x})^2} \\
\end{align*}\)
The quotient rule is not hard to understand. Just be careful.
The denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all over the denominator squared.
\(\displaystyle \frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{g(x)\frac{d}{dx}[f(x)]-f(x)\frac{d}{dx}[g(x)]}{[g(x)]^{2}}\)
\(\displaystyle g(x)\frac{d}{dx}[f(x)]\) will be 0 because the derivative of 144.44 is 0.
That just leaves you with:
\(\displaystyle \frac{-f(x)\frac{d}{dx}[g(x)]}{[g(x)]^{2}}\)
\(\displaystyle \frac{d}{dx}(1+\frac{383}{10000}e^{\frac{21x}{20}})\)
\(\displaystyle =\frac{8043e^{\frac{21x}{20}}}{200000}\)
\(\displaystyle \frac{(\frac{-3611}{25})\frac{8043e^{\frac{21x}{20}}}{200000}}{(1+\frac{383}{10000}e^{\frac{21x}{20}})^{2}\)
Oh, it's not a necessity, but try, if possible, expressing your terms as fractions; it just looks better than decimals.
The denominator times the derivative of the numerator minus the numerator times the derivative of the denominator all over the denominator squared.
\(\displaystyle \frac{d}{dx}[\frac{f(x)}{g(x)}]=\frac{g(x)\frac{d}{dx}[f(x)]-f(x)\frac{d}{dx}[g(x)]}{[g(x)]^{2}}\)
\(\displaystyle g(x)\frac{d}{dx}[f(x)]\) will be 0 because the derivative of 144.44 is 0.
That just leaves you with:
\(\displaystyle \frac{-f(x)\frac{d}{dx}[g(x)]}{[g(x)]^{2}}\)
\(\displaystyle \frac{d}{dx}(1+\frac{383}{10000}e^{\frac{21x}{20}})\)
\(\displaystyle =\frac{8043e^{\frac{21x}{20}}}{200000}\)
\(\displaystyle \frac{(\frac{-3611}{25})\frac{8043e^{\frac{21x}{20}}}{200000}}{(1+\frac{383}{10000}e^{\frac{21x}{20}})^{2}\)
Oh, it's not a necessity, but try, if possible, expressing your terms as fractions; it just looks better than decimals.
