please help to derive y=sin^{4 }(sec^{2 }3x) , I would like to know how to get the answer
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Originally Posted by MRina please help to derive y=sin^{4 }(sec^{2 }3x) , I would like to know how to get the answer $(f\circ g\circ h(x))'=(f'\circ g\circ h(x))(g'\circ h(x))(h'(x))$
Is this correct? d/dx= 24 sec^2(3x) tan(3x) cos(sec^2 (3x)) sin^3 (sec^2 (3x))
Originally Posted by MRina Is this correct? d/dx= 24 sec^2(3x) tan(3x) cos(sec^2 (3x)) sin^3 (sec^2 (3x)) Yes, good for you. Just as a matter of style I would encourage writing is "chain form".