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Llana [10]
1 year ago
10

ormula1" title="y = \frac{ \sqrt{3x} }{1 + e^{2x} } " alt="y = \frac{ \sqrt{3x} }{1 + e^{2x} } " align="absmiddle" class="latex-formula">
how would one differentiate this?
​
Mathematics
1 answer:
kumpel [21]1 year ago
7 0

~~~~~~~~y = \dfrac{\sqrt{3x}}{ 1+ e^{2x}}\\\\\\\implies \dfrac{dy}{dx} = \dfrac{d}{dx}  \left( \dfrac{\sqrt{3x}}{1+ e^{2x}} \right)\\\\\\~~~~~~~~~~~~=\dfrac{(1 + e^{2x} ) \dfrac{d}{dx} \left(\sqrt{3x} \right) - \left(\sqrt{3x} \right)\dfrac{d}{dx}(1+e^{2x})}{\left( 1+ e^{2x} \right)^2}~~~~~~~~~~~~;[\text{Quotient rule}]\\\\\\~~~~~~~~~~~~=\dfrac{(1+e^{2x}) \cdot \dfrac{1}{2\sqrt{3x}} \cdot 3-\left(\sqrt{3x} \right) \cdot 2 e^{2x}}{(1 + e^{2x})^2}~~~~~~~~~~~~~~~~~~~~;[\text{Chain rule}]\\\\\\

            =\dfrac{\dfrac{\sqrt 3 (1 + e^{2x})}{2\sqrt x} -2e^{2x} \sqrt{3x}}{(1+e^{2x})^2}\\\\\\=\dfrac{\tfrac{\sqrt 3(1 +e^{2x}) - 4x\sqrt 3 e^{2x}}{2\sqrt x}}{(1+e^{2x})^2}\\\\\\=\dfrac{\sqrt 3(1+e^{2x} -4xe^{2x})}{2\sqrt x(1 +e^{2x})^2}

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Vilka [71]
Multiply 4.51 (f) by d, the number of deliveries.

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5 0
3 years ago
Evaluate the function g(t)=3/2t-5/2,when t=5
klio [65]
G(t) = 3/2t - 5/2

g(5) = 3/2(5) - 5/2

g(5) = 15/2 - 5/2

g(5) = 10/2

g(5) = 5   is your answer

hope this helps
4 0
3 years ago
10
FromTheMoon [43]

The total amount shared by the friends illustrates mean and average, and the amount spent by each friend is $11.5

<h3>How to determine the amount?</h3>

The given parameters are:

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The amount spent by each friend is calculated using:

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So, we  have:

Amount = (76 + 16)/8

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Amount = 11.5

Hence, the amount spent by each friend is $11.5

Read more about mean and average at:

brainly.com/question/20118982

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8 0
2 years ago
Which table represents a linear function?​
amid [387]
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5 0
3 years ago
Daniel dives into a swimming pool. The height of his head is represented by the function h(t) = -8t2 - 28t + 60, where t is the
8_murik_8 [283]
So, h(t) is how far is Daniel's head from the surface of the water, namely the surface itself is when the height is nill, so his head is at the surface and the height of it is just 0, thus h(t)  = 0.  Namely, what is "t" when h(t) is 0?

\bf h(t)=-8t^2-28t+60\implies \stackrel{h(t)}{0}=-8t^2-28t+60&#10;\\\\\\&#10;0=-2t^2-7t+15\implies 2t^2+7t-15=0\implies (2t+3)(t-5)=0&#10;\\\\\\&#10;\begin{cases}&#10;2t+3=0\implies 2t=-3\implies &t=-\frac{3}{2}\\\\&#10;t-5=0\implies &\boxed{t=5}&#10;\end{cases}

clearly the seconds cannot be a negative unit, so is not -3/2.
6 0
3 years ago
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