Solve the problem below.

(15 Points)

expeople1 [14]

ANSWER 1

Note that,

$f(u)=tan^{-1}(u)$

is the same as

$f(u)=arctan(u)$

We apply the product rule.

$f(x)=x^2tan^{-1}(x)$

So we keep the second function and differentiate the first,plus we keep the first function and differentiate the second.

$f'(x)=(x^2)'tan^{-1}(x)+x^2(tan^{-1}(x))'$

Recall that,

If

$f(u)=tan^{-1}(u)$

Then,

$f'(u)=\frac{1}{1+u^2}} \times u'$

This implies that,

$f'(x)=2xtan^{-1}(x)+\frac{x^2}{x^2+1}$

ANSWER 2

We apply the product rule and the chain rules of differentiation here.

$f(x)=xsin^{-1}(1-x^2)$

$f'(x)=x'sin^{-1}(1-x^2)+x(sin^{-1}(1-x^2))'$

Recall that,

If

$f(u)=sin^{-1}(u)$

Then,

$f'(u)=\frac{1}{\sqrt{1-u^2}} \times u'$

This implies that,

$f'(x)=sin^{-1}(1-x^2)+x \times \frac{1}{\sqrt{1-(1-x^2)^2}}\times (-2x)$

$f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-(1-2x^2+x^4)}}$

$f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{1-1+2x^2-x^4}}$

$f'(x)=sin^{-1}(1-x^2)-\frac{2x^2}{\sqrt{2x^2-x^4}}$

5 0

4 years ago

Which of the following tables represents a function?

Karolina [17]

Answer: the second one

Step-by-step explanation:

6 0

3 years ago

PLEASE HELP I HAVE AN HOUR LEFT!!

Yuki888 [10]

The statement that correctly describes the horizontal asymptote of g(x) is:

Limit of g (x) as x approaches plus-or-minus infinity = 6, so g(x) has an asymptote at y = 6.

<h3>What are the asymptotes of a function f(x)?</h3>

The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.

The horizontal asymptote is the limit of f(x) as x goes to infinity, as long as this value is different of infinity.

In this problem, the function is:

$g(x) = \frac{42x^3 - 15}{7x^3 - 4x^2 - 3}$

The horizontal asymptote is given as follows:

$y = \lim_{x \rightarrow \infty} g(x) = \lim_{x \rightarrow \infty} \frac{42x^3 - 15}{7x^3 - 4x^2 - 3} = \lim_{x \rightarrow \infty} \frac{42x^3}{7x^3} = \lim_{x \rightarrow \infty} 6 = 6$

Hence the correct statement is:

Limit of g (x) as x approaches plus-or-minus infinity = 6, so g(x) has an asymptote at y = 6.

More can be learned about asymptotes at brainly.com/question/16948935

#SPJ1

5 0

2 years ago

A circular running track 1/4 is mile long. Elena runs on this track, completing each lap in 1/20 of an hour. Elena's running spe

Delicious77 [7]

Answer:

Elena's running speed is 5 miles/hour

Explanation:

The speed is defined as the covered distance per unit time

In the problem, we have:

The distance covered is the length (circumference) of the circular track which is given as $\frac{1}{4}$ of a mile.

We are also given that she completes each lap (she completes $\frac{1}{4}$ of a mile) in $\frac{1}{20}$ of an hour

To get her speed, we will divide the distance covered by the time taken to cover this distance

This is done as follows:

speed = $\frac{\frac{1}{4} }{\frac{1}{20} } = 5$ miles/hour

This means that:

Elena can run 5 miles each hour

Hope this helps :)

5 0

4 years ago

The circumference of the circle is 62.8cm what is the diameter use 3.14

Svetllana [295]

Answer:

20

Step-by-step explanation:

Divide 62.8 by 3.14

4 0

3 years ago