All categories

3 years ago

100 POINTS Evaluate the derivative of the function number 38

Mathematics

2 answers:

patriot [66]3 years ago

8 0

Answer:

1/ sqrt(1+ln^2(x)) * 1/(ln^2x +1) * 1/x

Step-by-step explanation:

f(x) = sin (tan^-1 (ln(x)))

u substitution

d/du (sin u) * du /dx

cos (u) * du/dx

Let u =(tan^-1 (ln(x))) du/dx =d/dx (tan^-1 (ln(x)))

v substitution

Let v = ln x dv/dx = 1/x

d/dv (tan ^-1 v) dv/dx

1/( v^2+1) * dv/dx

=1/(ln^2x +1) * 1/x

Substituting this back in for du/dx

cos (tan^-1 (ln(x)) * 1/(ln^2x +1) * 1/x

We know that cos (tan^-1 (a)) = 1/ sqrt(1+a^2)

cos (tan^-1 (ln(x)) * 1/(ln^2x +1) * 1/x

1/ sqrt(1+ln^2(x)) * 1/(ln^2x +1) * 1/x

ryzh [129]3 years ago

8 0

Answer:

cos((tan^-1)(lnx))/[x(1 + ln²x)]

Step-by-step explanation:

Sin((tan^-1)(lnx))

First differentiate the outer function 'sin'

Then inner functions

derivative of:

sin(x) is cos(x)

tan^-(x) = 1/(1+x²)

lnx = 1/x

cos((tan^-1)(lnx)) × [1/(1 + ln²x)] × 1/x

cos((tan^-1)(lnx))/[x(1 + ln²x)]

You might be interested in

There are 99 males and 121 females participating in a marathon. What percent of the participants are female?

ioda

To do this, we first want to see how many participants there are in total by adding both males and females.

$99+121=220$

Now we know that there were 220 total participants, but only 121 were females.

This allows us to set up the fraction 121/220= the percent female.

$\frac{121}{220} =0.55$

or 55%

8 0

3 years ago

Find the slope and put the answer in simplest form

vlabodo [156]

Answer:

Slope is 6

Step-by-step explanation:

4 0

3 years ago

The function f(x) varies inversely with x and f(x) = 0.9 when x = 0.5 what is f(x) when x = 1.5 ?

marin [14]

When x = 0.5, the answer is 0.45
When x = 1.5, the answer is 1.35

You're welcome :)

4 0

3 years ago

Read 2 more answers

I need the domain range and function. With an explanation

erastovalidia [21]

<h3>Answers:</h3>

Problem 1

Domain = $-3 < x \le 3$ , interval notation (-3, 3]
Range = $-3 \le y < 3$ , interval notation [-3, 3)
Is it a function? Yes

Problem 2

Domain = $x \ge -2$ , interval notation $[-2, \infty)$
Range = All real numbers, interval notation $(-\infty, \infty)$
Is it a function? No

Problem 3

Domain = $-4 \le x < 3$ , interval notation [-4, 3)
Range = $-4 < y \le 3$ , interval notation (-4, 3]
Is it a function? Yes

Problem 4

Domain = All real numbers, interval notation $(-\infty, \infty)$
Range = $y \le 4$ , interval notation $(-\infty, 4]$
Is it a function? Yes

==================================================

Explanations:

The left most point is when x = -3, and we are not including this value due to the open hole. The other endpoint is included because it is a filled in circle. The domain is therefore $-3 < x \le 3$ which in interval notation is (-3, 3]. We have the curved parenthesis meaning "exclude endpoint" and the square bracket says "include endpoint". The range is a similar story but we're looking at the smallest and largest y values. Though be careful about which endpoint is open/closed. We have a function because it passes the vertical line test.
The smallest x value is x = -2. There is no largest x value because the arrows say to go on forever to the right. We can say the domain is $x \ge -2$ which in interval notation is $[-2, \infty)$ . The range is $(-\infty, \infty)$ to indicate "all real numbers". This graph fails the vertical line test, so it is not a function. The vertical line test is where we check to see if we can pass a vertical line through more than one point on the curve. In this case, such a thing is possible which is why it fails the test.
This is the same idea as problem 1, though note the endpoints are flipped in terms of which has an open circle and which doesn't. It is not possible to draw a single vertical line to have it pass through more than one point on the curve, so it passes the vertical line test and we have a function.
This is a function because it passes the vertical line test. The domain is the set of all real numbers due to the arrows in both directions. Any x value is a possible input. The range is $y \le 4$ which is the same as saying $(-\infty, 4]$ in interval notation. This is because y = 4 is the largest y value possible. There is no smallest y value due to the arrows.

7 0

4 years ago

4x² – 3x²<br> Please help <br> i promise ill study next time

irakobra [83]

Answer:

1 $x^{2}$

Hope that helps!

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers