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प्रश्न
Find `"dy"/"dx"` ; if x = sin3θ , y = cos3θ
योग
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उत्तर
x = sin3 θ
differentlating w.r.t. θ
`"dy"/("d" theta) = 3 "sin"^2 theta . "cos" theta`
Y = cos3θ
Differentiating w.r.t. θ
`"dy"/("d" theta) = 3"cos"^2 theta (-"sin" theta)`
= -3 cos2 θ . sin θ
`"dy"/"dx" = ("dy"/("d"theta))/("dx"/("d" theta)) = (-3"cos"^2 theta . "sin" theta)/(3"sin"^2 theta . "cos" theta) = ("- cos" theta)/("sin" theta) = - "cot" theta`
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