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Dmitriy789 [7]
2 years ago
10

Find the derivative of

" align="absmiddle" class="latex-formula"> by 1st principle of derivative.
Mathematics
1 answer:
sladkih [1.3K]2 years ago
6 0

\huge{\color{magenta}{\fbox{\textsf{\textbf{Answer}}}}}

\frak {\huge{ \frac{1}{1 +  {x}^{2} } }}

Step-by-step explanation:

\sf let \: f(x) =  { \tan }^{ - 1} x \\  \\  \sf f(x + h) =  { \tan}^{ - 1} (x + h)

\sf f'(x) =  \frac{f(x+h)  - f(x) }{h}

\sf \implies \lim_{  h \to 0  } \frac{ { \tan }^{ - 1}(x + h) -  { \tan}^{ - 1}x  }{h}  \\  \\  \\  \sf  \implies  \lim_ {h \to 0}    \frac{  { \tan}^{ - 1} \frac{x + h - x}{1 + (x + h)x} }{h}

By using

\sf { \tan}^{ - 1} x -  { \tan}^{ - 1} y   = \\   \sf { \tan}^{ - 1}  \frac{x - y}{1 + xy} formula

\sf  \implies  \large \lim_{h \to0 }   \frac{  { \tan}^{ - 1}  \frac{h}{1 + hx +  {x}^{2} } }{h}  \\  \\  \\  \sf  \implies   \large{\lim_{h \to0}   } \frac{ { \tan}^{ - 1}  \frac{h}{1 + hx +  {x}^{2} } }{ \frac{h}{1 + hx  +  {x}^{2} }  \times (1 + hx +  {x}^{2} )}  \\  \\  \\  \sf  \implies \large  \lim_{h \to0} \frac{ { \tan}^{ - 1} \frac{h}{1 + hx +  {x}^{2} }  }{ \frac{h}{1 + hx +  {x}^{2} } }  +  \lim_{h \to0} \frac{1}{1 + hx +  {x}^{2} }

<u>Now</u><u> </u><u>putting</u><u> </u><u>the</u><u> </u><u>value</u><u> </u><u>of</u><u> </u><u>h</u><u> </u><u>=</u><u> </u><u>0</u>

<u>\sf  \large  \implies 0 +  \frac{1}{1 + 0 +  {x}^{2} }  \\  \\  \\  \purple{ \boxed  { \implies  \frac{1}{1 +  {x}^{2} } }}</u>

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WILL GIVE BRAINIEST ANSWER!!! TUVW has vertices T(1,-3),U(3,-1),V(6,-4),and W(4,-6).Determine if quadrilateral TUVW is a paralle
Arte-miy333 [17]
Plotting the 4 points and doing visual inspection will tell us if the quadrilateral is a parallelogram. We can also do analytical techniques which are to determine the slope and the distance between the points. Doing either of these two will give us
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ladessa [460]

The linear equation in standard form is:

6x - 7y = -11

Which is the last option.

<h3>How to get the equation of the line?</h3>

The line in slope-intercept form is written as:

y = a*x + b

We can see that the line passes through the points (-3, -1) and (1/2, 2), then the slope is:

a = \frac{2 - (-1)}{1/2 - (-3)}  = \frac{3}{3.5}  = \frac{6}{7}

Then we can write:

y = (6/7)*x + b

To find the value of b, we use the first point. It means that when x = -3, the value of y is -1, then we get:

-1 = (6/7)*-3 + b

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11/7 = b

Then the equation is:

y = (6/7)*x + 11/7

If we multiply both sides by 7 we get:

7y = 6x + 11

Now we move the term with "x" to the left:

7y - 6x = 11

That is the line in standard form.

If we multiply both sides by -1, we get the last option:

6x - 7y = -11

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

8 0
2 years ago
Read 2 more answers
A line passes through the points (–5, 2) and (10, –1). Which is the equation of the line?
LenKa [72]
We have y - 2 = m(x + 5), where m = (-1 - 2)/(10 + 5) = -3/15 = -1/5;
Then y - 2 = (-1/5)(x + 5);
5y - 10 = (-1)(x + 5);
5y - 10 = -x - 5;
x + 5y - 5 = 0 is the equation of the line.
3 0
3 years ago
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