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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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The coordinates of point B' are (-3, 6).

<h2>Given that</h2>

Triangle ABC has vertices A(4, 5), B(0, 5), and C(3, –1).

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<h3>We have to determine</h3>

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Triangle ABC has vertices A(4, 5), B(0, 5), and C(3, –1).

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