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3 years ago

How do you calculate the derivative of y with respect to x on the equation x^3y+3xy^3=x+y?

1 answer:

3 0

$\bf x^3y+3xy^3=x+y\\\\
-----------------------------\\\\
\left[ 3x^2y+x^3\frac{dy}{dx} \right]+3\left[ 1\cdot y^3+x\cdot 3y^2\frac{dy}{dx} \right]=1+\frac{dy}{dx}
\\\\\\
3x^2y+x^3\frac{dy}{dx}+3y^3+9xy^2\frac{dy}{dx}-\frac{dy}{dx}=1\impliedby \textit{common factor}
\\\\\\
\cfrac{dy}{dx}(x^3+9xy^2-1)+3x^2y+3y^3=1
\\\\\\
\cfrac{dy}{dx}(x^3+9xy^2-1)=1-3x^2y-3y^3
\\\\\\
\cfrac{dy}{dx}=\cfrac{1-3x^2y-3y^3}{x^3+9xy^2-1}$

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Step-by-step explanation:

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Melinda ran 8,791 meters on Monday and 2,899 meters on Tuesday. How many more meters did she run on Monday than on Tuesday?

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2 years ago

Suppose that the population of the scores of all high school seniors who took the SAT Math (SAT-M) test this year follows a Norm

Vlad1618 [11]

Answer:

Confidence = $1-\alpha=1-0.01=0.99$

And then the confidence level would be given by 99%

Step-by-step explanation:

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Assuming the X follows a normal distribution

$X \sim N(\mu, \sigma)$

The distribution for the sample mean is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\bar x =512$ represent the sample mean

$\sigma=100$ represent the population standard deviation

n= 100 sample size selected.

The confidence interval is given by this formula:

$\bar X \pm z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$ (1)

The marginof error for this case is given by Me=25.76. And we know that the formula for the margin of error is given by:

$Me=z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$

$25.76=z_{\alpha/2} \frac{100}{\sqrt{100}}$

And we can find the critical value $z_{\alpha/2}$ like this:

$z_{\alpha/2}=\frac{25.76(\sqrt{100})}{100}=2.576$

And we know that on the right tail of the z score =2.576 we have $\alpha/2$ of the total area. We can find the area on the right of the z score using this excel code:

"=1-NORM.DIST(2.576,0,1,TRUE)" or using a table of the normal standard distribution, and we got 0.004998= $\alpha/2$ , so then $\alpha=0.00498*2=0.009995 \approx 0.01$ , and then we can find the confidence like this: