Question

Y=log(log sinx) find dy/dx

Answer 1

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➡Find Here.

♠ dy/dx , if y = log(log sinx).

➡Differenciate w.r.to X.

=> dy/dx = d/dx [ log(log sinx) ].

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=> dy/dx = d/dx [ log(log sinx) ].

=> dy/dx = [ {1/log sinx} (d( log sinx )/dx].

=> dy/dx = [ {1/(log sinx)} (1/sinx) d sinx /dx ].

=> dy/d. = [ Cos x /{sin x (log sinx )}].

=> dy/dx = Cot x /log sinx .

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➡Important Point .

♠ d sinx/dx = Cos x .

♠ d log x /dx = 1/x.

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