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If `y=(1)/(logcosx)`, find `(dy)/(dx).` |
Answer» Given : `y=(log cos x)^(-1).` Putting `cosx=t` and `log cos x=logt=u`, we get `y=u^(-1)=(1)/(u),u=log t and t = cos x` `rArr(dy)/(dx)=(-1)/(u^(2)),(du)/(dt)=(1)/(t) and (dt)/(dx)=-sinx` `rArr(dy)/(dx)=((dy)/(dx)xx(du)/(dt)xx(dt)/(dx))` `={(-1)/(u^(2))xx(1)/(t)xx(-sinx)}={(1)/((log cos x)^(2)).(1)/(cos x).sinx}` `=(tanx)/((log cos x)^(2))` |
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