Question

dy/dx=?

Answer 1

\[\LARGE y=x^{\cos^{-1}x} \]

Answer 2

my attempt:
\[\LARGE \log y=\log x^{\cos^{-1}x}\]

Answer 3

\[\LARGE \frac{1}{y} \times \frac{dy}{dx}= \frac{1}{x^{\cos^{-1}x}} \times \frac{1}{-\sqrt{1-\cos^{-2}x}} \]
something like this perhaps?

Answer 4

@hartnn

Answer 5

log x^a = a log x
then use product rule.

Answer 6

I would use log(x^a)= alog(x)

Answer 7

dam confused :| okay

Answer 8

\(\LARGE \log y=\log x^{\cos^{-1}x} =\cos^{-1}x\log x \)

Answer 9

product rule on right side.

Answer 10

\[\frac{d}{dy}logy=d{\cos^{-1}xlogx}\]

Answer 11

\[\LARGE \frac{1}{y} \times \frac{dy}{dx}=\cos^{-1}x \frac{1}{x}+\frac{\log}{-\sqrt {1-\cos^{-2}x}} \]
this?

Answer 12

logx*

Answer 13

d/dx cos^-1 of x =.... ?

Answer 14

\[\frac{-1}{\sqrt{1-x^{2}}}\]

Answer 15

\(\LARGE \frac{1}{y} \times \frac{dy}{dx}=\cos^{-1}x \frac{1}{x}+\frac{\log x}{-\sqrt {1-x^2}}\)

Answer 16

okay sorry :S thanks :D (again)

Answer 17

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