Question

Ify= (sinx + cosx)² then find dy/dx
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Answer 1

Take log on both sides,

log y = log (sin x)^(cos x)

=> log y = cos(x)*log(sin x) [Property of log, log x^n = n log x]

=> (1/y)(dy/dx) = [log(sin x)*(d(cos x)/dx)] + [ cos(x)d(log(sinx))/dx]

=>dy/dx = y[log(sin x)*(-sin x) + cosx*(1/sin x)*cos(x)]

We have y = (sin(x))^cos(x)

dy/dx = [(sin(x))^cos(x)][log(sin x)*(-sin x) + cosx (1/sin x)*cos(x)]

dy/dx = (sin x)^cos(x) [ log (sin x)^-sin(x) + cot(x)cos(x)]

Answer 2

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