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प्रश्न
If x = cos2 θ and y = cot θ then find `dy/dx at θ=pi/4`
उत्तर १
`x=cos^2θ and y=cot θ`
`(dx)/(dθ)=d/(dθ) (cos^2θ)`
`dx/(dθ)=-2cosθ sin θ`
`dy/dθ=-cosec^2θ`
`dy/dx=dy/(dθ)/dx/(dθ)`
= `(-cosec^2θ)/(-2cosθ sinθ)`
=`1/(2sin^3 θ cos θ)`
=`(1/2sin^3θ cos θ)θ=pi/4`
`(dy/dx)_θ=pi/4`
=`1/2(1/sqrt2)^3 1/sqrt2`
=`1/(2 1/4)=2`
उत्तर २
`x=cos^2θ and y=cot θ`
`(dx)/(dθ)=2 cosθ (-sinθ )`
`dx/(dθ)=-2cosθ sin θ`
y = cotθ
`dy/dθ=-cosec^2θ`
`dy/dx=(dy/(dθ))/(dx/(dθ))`
= `(-cosec^2θ)/(-2cosθ sinθ)`
`((dy)/(dx))_(0=π/4) = (cosec^2 π/4)/(2.sin π/4. cos π/4) `
= `2/(2 xx 1/sqrt2 xx 1/sqrt2`
= 2
उत्तर ३
`x=cos^2θ and y=cot θ`
`(dx)/(dθ)=2 cosθ (-sinθ )`
`dx/(dθ)=-2cosθ sin θ`
y = cotθ
`dy/dθ=-cosec^2θ`
`dy/dx=(dy/(dθ))/(dx/(dθ))`
= `(-cosec^2θ)/(-2cosθ sinθ)`
`((dy)/(dx))_(0=π/4) = (cosec^2 π/4)/(2.sin π/4. cos π/4) `
= `2/(2 xx 1/sqrt2 xx 1/sqrt2`
= 2
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