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Lena [83]
2 years ago
12

How to find the derivative???

Mathematics
1 answer:
kakasveta [241]2 years ago
4 0

Answer:

y'=  \dfrac{2}{3}e^{2x + e^{2x}/3}

Step-by-step explanation:

We have e^{2x} = \ln(y^3) and we want the implicit differentiation \dfrac{dy}{dx}f(x) = y'

Therefore, we want to separate the y from the equation.

\ln(y^3) =e^{2x} \implies 3 \ln(y) = e^{2x}  \implies  \ln(y)  = \dfrac{e^{2x} }{3}

\implies e^{\ln(y)} = e^{e^{2x}/3} \implies \boxed{y =  e^{e^{2x}/3}}

Now we can calculate the derivative. By the chain rule,

y' = \dfrac{d}{dx}\left(e^{e^{2x}/3}\right) = e^{e^{2x}/3}\dfrac{d}{dx}\left(\dfrac{e^{2x}}{3}\right)

Now

\dfrac{d}{dx}\left(\dfrac{e^{2x}}{3}\right) =\dfrac{1}{3}\dfrac{d}{dx}\left(e^{2x}\right) = \dfrac{2}{3}e^{2x}

Therefore

y' = e^{e^{2x}/3}\cdot \dfrac{2}{3}e^{2x} = \dfrac{2}{3}e^{2x + e^{2x}/3}

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PLEASE HELP A flight across the US takes longer east to west then it does west to east. This is due to the plane having a headwi
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To solve our questions, we are going to use the kinematic equation for distance: d=vt
where
d is distance 
v is speed  
t is time 

1. Let v_{w} be the speed of the wind, t_{w} be time of the westward trip, and t_{e} the time of the eastward trip. We know from our problem that the distance between the cities is 2,400 miles, so d=2400. We also know that the speed of the plane is 450 mi/hr, so v=450. Now we can use our equation the relate the unknown quantities with the quantities that we know:

<span>Going westward:
The plane is flying against the wind, so we need to subtract the speed of the wind form the speed of the plane:
</span>d=vt
2400=(450-v_{w})t_{w}

Going eastward:
The plane is flying with the wind, so we need to add the speed of the wind to the speed of the plane:
d=vt
2400=(450+v_{w})t_{e}

We can conclude that you should complete the chart as follows:
Going westward -Distance: 2400 Rate:450-v_w Time:t_w
Going eastward -Distance: 2400 Rate:450+v_w Time:t_e

2. Notice that we already have to equations:
Going westward: 2400=(450-v_{w})t_{w} equation(1)
Going eastward: 2400=(450+v_{w})t_{e} equation (2)

Let t_{t} be the time of the round trip. We know from our problem that the round trip takes 11 hours, so t_{t}=11, but we also know that the time round trip is the time of the westward trip plus the time of the eastward trip, so t_{t}=t_w+t_e. Using this equation we can express t_w in terms of t_e:
t_{t}=t_w+t_e
11=t_w+t_e equation
t_w=11-t_e equation (3)
Now, we can replace equation (3) in equation (1) to create a system of equations with two unknowns: 
2400=(450-v_{w})t_{w}
2400=(450-v_{w})(11-t_e) 

We can conclude that the system of equations that represent the situation if the round trip takes 11 hours is:
2400=(450-v_{w})(11-t_e) equation (1)
2400=(450+v_{w})t_{e} equation (2)

3. Lets solve our system of equations to find the speed of the wind: 
2400=(450-v_{w})(11-t_e) equation (1)
2400=(450+v_{w})t_{e} equation (2)

Step 1. Solve for t_{e} in equation (2)
2400=(450+v_{w})t_{e}
t_{e}= \frac{2400}{450+v_{w}} equation (3)

Step 2. Replace equation (3) in equation (1) and solve for v_w:
2400=(450-v_{w})(11-t_e)
2400=(450-v_{w})(11-\frac{2400}{450+v_{w}} )
2400=(450-v_{w})( \frac{4950+11v_w-2400}{450+v_{w}} )
2400=(450-v_{w})( \frac{255011v_w}{450+v_{w}} )
2400= \frac{1147500+4950v_w-2550v_w-11(v_w)^2}{450+v_{w}}
2400(450+v_{w})=1147500+2400v_w-11(v_w)^2
1080000+2400v_w=1147500+2400v_w-11(v_w)^2
(11v_w)^2-67500=0
11(v_w)^2=67500
(v_w)^2= \frac{67500}{11}
v_w= \sqrt{\frac{67500}{11}}
v_w=78

We can conclude that the speed of the wind is 78 mi/hr.
6 0
3 years ago
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