I need help with this math question please

Mathematics

1 answer:

Gemiola [76]3 years ago

8 0

Answer:

yes D. is correct

Step-by-step explanation:

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Find the derivative of <img src="https://tex.z-dn.net/?f=tan%5E%7B-1%7D%20x" id="TexFormula1" title="tan^{-1} x" alt="tan^{-1} x

sladkih [1.3K]

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

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

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

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

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