Question

find dy÷dx if you =sin[ cos ( tan x) ]

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Answer 1

Step-by-step explanation:

1/ydy/dx=[tanx(d/dxlog sinx)+log sin x(d/dxtanx)]

1/ydy/dx=[tanx 1/sinx (d/dxsin x)+log sinx sec^2x]

1/y dy/dx=[tanx 1/sinx.cosx+logsinx sec^2x]

1/y dy/dx=[tanx.cosx/sinx+logsinx sec^2x]

1/y dy/dx=[sinx/cosx.cosx/sinx +logsinx sec^2x]

1/y dy/dx=[1+log sinx sec^2x]

dy/dx=y[1+log sinx sec^2x]

dy/dx=sinx^tanx[1+logsinx sec^2x]