find dy/dx, if y=log sin x^2
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dy/dx. =( 1/sinx^2 ) cosx^2. 2x
neha287:
plz fir se
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1
y=log sin x^2= dy/dx=1/sin x^2(cos x^2)(2x)=2x cot x^2
plz fir se btao
pura solution
dy/dx = d(log sinx^2)/dx × d(sin x^2)/dx × d(x^2)/dx
dy/dx= (1/sinx^2) × (cosx^2) × (2x) = 2x cotx^2 using:-) sinx/cosx =cot x...would be the whole solution...
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