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missed my lecture and need help with this derivative, I tried looking at online resources but I just got confused so if someone

Zigmanuir [339]

Derivative with respect to x: $12x^2y^2-\frac{z^3}{x}+16x^{15}$

Derivative with respect to y: $8x^3y-4e^zy^3+4$

Derivative with respect to z: $-e^z y^4 +\frac{3z^2}{x^2}$

Step-by-step explanation:

We can calculate three derivatives for this function: the derivative with respect to x, the derivative with respect to y, and the derivative with respect to z.

We start with the derivative with respect to x:

$\frac{df}{dx}=4(3x^{3-1})y^2 -0 +z^3(-x^{-1})+0-16x^{16-1}=\\=12x^2y^2-\frac{z^3}{x}+16x^{15}$

Then we calculate the derivative with respect to y:

$\frac{df}{dy}=4x^3(2y^{2-1})-e^z(4y^{4-1})+0+4y^{1-1}=8x^3y-4e^zy^3+4$

Finally, we calculate the derivative with respect to z:

$\frac{df}{dz}=0-e^z y^4 +\frac{3z^{3-1}}{x^2}+0-0=-e^z y^4 +\frac{3z^2}{x^2}$

Learn more about differentiation:

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3 years ago

In a clinical trial of 268 subjects treated with a drug, 11% of the subjects reported dizziness. The margin of error is plus o

Mariulka [41]

Answer:

$\hat p =0.11$ represent the proportion estimated of subjects reported with dizziness

n = 268 represent the random sample selected

$\hat q = 1-\hat p = 1-0.11= 0.89$ represent the proportion of subjects No reported with dizziness

E= 0.04 = 4% represent the margin of error for the confidence interval

$\alpha= 1-0.90 =0.1$ and this value represent the significance level of the test or the probability of error type I

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$