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differentiate and simplify: x=tan(x+y)
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\[x+y=\tan^{-1} x\] diff. with respect to x can you complete it?
\[\frac{ d }{dx }\left( \tan^{-1} x \right)=\frac{ 1 }{ 1+x^2 }\]
do you know ,how to solve it?
\[1+\frac{ dy }{dx }=\frac{ 1 }{1+x^2}\] find \[\frac{ dy }{ dx }.\]
if you have not leaned inverse trigonometric functions. you can proceed directly
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diff. w.r.t. x \[1=\sec ^2\left( x+y \right)*\frac{ d }{ dx }\left( x+y \right)\]
that is the first step i know. not inverse. the answer is cos^2(x+y)-1
i just can't get there
what is \[\frac{ d }{ dx }\left( x+y \right)=?\]
\[=1+\frac{ dy }{dx }\]
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thanks, but I need the steps.
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