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Artemon [7]
3 years ago
5

Use implicit differentiation to solve that the derivative

Mathematics
1 answer:
Len [333]3 years ago
7 0

Given

<em>e</em> ˣʸ = sec(<em>x</em> ²)

take the derivative of both sides:

d/d<em>x</em> [<em>e</em> ˣʸ] = d/d<em>x</em> [sec(<em>x</em> ²)]

Use the chain rule:

<em>e</em> ˣʸ d/d<em>x</em> [<em>xy</em>] = sec(<em>x</em> ²) tan(<em>x</em> ²) d/d<em>x</em> [<em>x</em> ²]

Use the product rule on the left, and the power rule on the right:

<em>e</em> ˣʸ (<em>x</em> d<em>y</em>/d<em>x</em> + <em>y</em>) = sec(<em>x</em> ²) tan(<em>x</em> ²) (2<em>x</em>)

Solve for d<em>y</em>/d<em>x</em> :

<em>e</em> ˣʸ (<em>x</em> d<em>y</em>/d<em>x</em> + <em>y</em>) = 2<em>x</em> sec(<em>x</em> ²) tan(<em>x</em> ²)

<em>x</em> d<em>y</em>/d<em>x</em> + <em>y</em> = 2<em>x</em> <em>e</em> ⁻ˣʸ sec(<em>x</em> ²) tan(<em>x</em> ²)

<em>x</em> d<em>y</em>/d<em>x</em> = 2<em>x</em> <em>e</em> ⁻ˣʸ sec(<em>x</em> ²) tan(<em>x</em> ²) - <em>y</em>

d<em>y</em>/d<em>x</em> = 2<em>e</em> ⁻ˣʸ sec(<em>x</em> ²) tan(<em>x</em> ²) - <em>y</em>/<em>x</em>

Since <em>e</em> ˣʸ = sec(<em>x</em> ²), we simplify further to get

d<em>y</em>/d<em>x</em> = 2 tan(<em>x</em> ²) - <em>y</em>/<em>x</em>

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Complete Question

A random sample of 300 circuits generated 13 defectives. a. Use the data to test

                            H_o : p = 0.05

Versus

 

                          H_1 : p \ne 0.05

Use α = 0.05. Find the P-value for the test

   

Answer:

The  p-value is  p-value =  0.5949        

Step-by-step explanation:

From the question we are told that

  The sample size is  n = 300

    The number of defective circuits is  k = 13

Generally the sample proportion of defective circuits is mathematically represented as

        \^ p = \frac{k}{n}

=>     \^ p = \frac{13}{300}

=>     \^ p = 0.0433

Generally the standard Error is mathematically represented as

       SE  = \sqrt{\frac{p(1- p)}{n} }

=>     SE  = \sqrt{\frac{0.05(1- 0.05)}{300} }

=>     SE  = 0.0126

Generally the test statistics is mathematically represented as

       z =  \frac{\^ p - p }{SE}

=>     z =  \frac{0.0433 - 0.05 }{0.0126}

=>     z =  -0.5317

From the z table  the area under the normal curve to the left  corresponding to  -0.5317  is

         (P < -0.5317 ) = 0.29747

Generally the p-value is mathematically represented as

       p-value =  2 * P(Z <  -0.5317 )

=>     p-value =  2 * 0.29747

=>     p-value =  0.5949        

     

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Answer:

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Step-by-step explanation:

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y=k(1/X^3)

1=k(1/6^3)

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