Find dy/dx:: sin xy + cos …
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Payal Singh 6 years, 10 months ago
sin xy + cos xy = 1
Differentiate with respect to x, we get
cos xy ( y + x{tex}dy\over dx{/tex}) - sin xy (y+x{tex}dy\over dx{/tex}) = 0
=> cos xy ( y + x{tex}dy\over dx{/tex}) = sin xy (y+x{tex}dy\over dx{/tex})
=> tan xy = 1
Differentiate with respect to x,
=> sec2 xy (y+ x{tex}dy\over dx{/tex}) = 0
=> y+ x {tex}dy\over dx{/tex}= 0
=> {tex}dy\over dx{/tex} = {tex}-y\over x{/tex}
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