If y=sin u, u=v^2 and v=4(x), then what is dy/dx?

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Answer 1 · 2018-01-09T04:42:19

#dy/dx=cos(16x^2)*32x#

#y=sin(u)#

#u=v^2#, #v=4x#, so #u=16x^2#

#y=sin(16x^2)#

Let #a=16x^2#

#dy/dx=(sin(a))'*a'#

#dy/dx=cos(a)*32x#

Put back #a=16x^2#,

#dy/dx=cos(16x^2)*32x#