Question

x=(cost)ᵗ y=(sint)ᵗ 0 < t < π/2,Find dy/dx for the given function y wherever defined

Answer 1

x = (cost)^t = e^{t.logcost}

differentiate with respect to t,

dx/dt = d[e^{t.logcost}]/dt

= e^{t logcost} × d[t.logcost]/dt

= (cost)^t × [t × d(logcost)/dt + logcost × dt/dt ]

= (cost)^t × [t × (-sint)/cost + logcost × 1]

= (cost)^t [ -t.tant + logcost]

= x[-t.tant + logcost]

similarly, y = (sint)^t = e^{t.logsint}

differentiate with respect to t,
dy/dt = e^{t.logsint} × d[t logsint]/dt

= (sint)^t [t × d(logsint)/dt + logsint × dt/dt ]

= (sint)^t [ t × cost/sint + logsint]

= (sint)^t [t.cott + logsint]

= y[t.cott + logsint]

now, dy/dx = {dy/dt}/{dx/dt}

dy/dx = y[t.cott + logsint]/x[-t.tant +logcost]

Answer 2

HELLO DEAR,




GIVEN:-
x = (cost)^t

taking log both side,

logx = t log(cost)

differentiating w.r.t t

(1/x)*dx/dt = t.d{log(cost)}/dt + log(cosx).d(t)/dt

(1/x)*dx/dt = -t.tant + log(cosx)

dx/dt = (cost)^t [-t.tant + log(cosx)]-------( 1 )

AND,
Y = (sint)^t

taking log both side,

logy = t.log(sint)

differentiating w.r.t t

(1/y)*dy/dt = tcot t + log(sinx)

dy/dt = (sint)^t[tcot t + log(sinx)]------( 2 )

divide-------( 1 ) by --------( 2 )

dy/dx = {dy/dt}/{dx/dt}

dy/dx = [(sint)^t{tcot t + log(sinx)}]/[(cost)^t{-t.tan t + log(cosx)}]

dy/dx = (tan t)^t.[t cot t + log(sinx)] / [-t tan t + log(cosx)]



I HOPE ITS HELP YOU DEAR,
THANKS