Asked by help me pls
Prove that if y = cotx then dy/dx = - (cscx)^2. Hint: cotx = cosx/sinx
Answers
Answered by
mathhelper
y = cotx
y = cosx/sinx
dy/dx = (sinx(-sinx) - cosx(cosx) )/sin^2 x
this assumes you know how to take derivatives of sinx and cosx
= -(sin^2 x + cos^2 x)/sin^2 x
= -1/sin^2 x
= -csc^2 x or - (cscx)^2
y = cosx/sinx
dy/dx = (sinx(-sinx) - cosx(cosx) )/sin^2 x
this assumes you know how to take derivatives of sinx and cosx
= -(sin^2 x + cos^2 x)/sin^2 x
= -1/sin^2 x
= -csc^2 x or - (cscx)^2
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