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3241004551 [841]
2 years ago
9

Y=arccos(1/x) Please help me do them all! I don’t know derivatives :(

Mathematics
1 answer:
Stells [14]2 years ago
4 0

Answer:

f(x) =  {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x  \rightarrow \: x = sec \: y\\  \frac{dx}{dx}  =  \frac{d(sec \: y)}{dx}  \\ 1 = \frac{d(sec \: y)}{dx} \times  \frac{dy}{dy}  \\ 1 = \frac{d(sec \: y)}{dy} \times  \frac{dy}{dx}  \\1 = tan \: y.sec \: y. \frac{dy}{dx}  \\ \frac{dy}{dx} =  \frac{1}{tan \: y.sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ \sqrt{( {sec}^{2}   \: y - 1}) .sec \: y}  \\ \frac{dy}{dx} =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\   \therefore  \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx}  =  \frac{1}{ |5x |  \sqrt{25 {x }^{2}  - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx}  =  \frac{1}{ |x |  \sqrt{ {x }^{2}  - 1} }

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