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If X = Cos2 θ and Y = Cot θ Then Find D Y D X a T θ = π 4

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प्रश्न

If x = cos2 θ and y = cot θ then find `dy/dx  at  θ=pi/4` 

योग
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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`

 

shaalaa.com

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 

shaalaa.com

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 

shaalaa.com
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2017-2018 (March)

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