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Differentiate y= exp^(3x^2)+x; f(x)= 100/2+9exp^(3x)
- Thread starter ladyazpy
- Start date
D
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First you need to define - what the question is.
I suppose you have been asked to find first derivative of the given functions.
The first one looks good.
#2
f(x) = 100/{2+9exp^(3x) }^(2) = 100 * {2+9exp^(3x) }^(-2)
f'(x) = 100 * (-2) * {9*3*exp(3x)} *{2+9exp^(3x) }^(-3)
I suppose you have been asked to find first derivative of the given functions.
The first one looks good.
#2
f(x) = 100/{2+9exp^(3x) }^(2) = 100 * {2+9exp^(3x) }^(-2)
f'(x) = 100 * (-2) * {9*3*exp(3x)} *{2+9exp^(3x) }^(-3)
Re: please check my work
Hello, ladyazpy!
Assuming that the second problem is: \(\displaystyle \L\:f(x)\;=\;\frac{100}{2\,+\,9e^{3x}}\)
. . we have: \(\displaystyle \L\:f(x) \;=\;100\left(2\,+\,9e^{3x}\right)^{-1}\)
Then: \(\displaystyle \L\:f'(x) \;=\;-100\left(2\,+\,9e^{3x}\right)^{-2}\cdot\,9e^{3x}\,\cdot\,3 \;=\;\frac{-2700e^{3x}}{(2\,+\,9e^{3x})^2}\)
Hello, ladyazpy!
Assuming that the second problem is: \(\displaystyle \L\:f(x)\;=\;\frac{100}{2\,+\,9e^{3x}}\)
. . we have: \(\displaystyle \L\:f(x) \;=\;100\left(2\,+\,9e^{3x}\right)^{-1}\)
Then: \(\displaystyle \L\:f'(x) \;=\;-100\left(2\,+\,9e^{3x}\right)^{-2}\cdot\,9e^{3x}\,\cdot\,3 \;=\;\frac{-2700e^{3x}}{(2\,+\,9e^{3x})^2}\)