find dy÷dx of (cos(x))^y = (sin(y))^x
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Take log both the side , u’ll get , y logcosx = x logsinyDifferentiate both the sides , y’logcosx –y/cosx * -sinx = logsiny +x/siny * coty *y’y’[logcosx-xcoty] = logsiny +ycotxy’=logsiny+ ycotx/ logcosx-xcoty y' = dy/dx.I hope this answer is helpful to you
Harshmourya1:
would you plz answer me dy÷dx (logcosx)
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