Question 16 (Or 2nd) - CBSE Class 12 Sample Paper for 2019 Boards
Last updated at Oct. 1, 2019 by Teachoo

Question 16 (OR 2
nd
question)

If y = log (1 + 2t
^{
2
}
+ t
^{
4
}
), x = tan
^{
-1
}
t, find d
^{
2
}
y/dx
^{
2
}

^{
}

^{
}

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Transcript

Question 16 (OR 2nd question) If y = log〖(1+2𝑡^2+𝑡^4)〗, x = tan^(−1)𝑡, find (𝑑^2 𝑦)/(𝑑𝑥^2 )
Finding 𝒅𝒚/𝒅𝒕
y = log〖(1+2𝑡^2+𝑡^4)〗
𝑑𝑦/𝑑𝑡 = 1/((1 + 2𝑡^2 + 𝑡^4 ) ) × 𝑑(1 + 2𝑡^2 + 𝑡^4 )/𝑑𝑡
𝑑𝑦/𝑑𝑡 = ((4𝑡^3 + 4𝑡))/((1 + 2𝑡^2 + 𝑡^4 ) )
𝑑𝑦/𝑑𝑡 = (4𝑡(𝑡^2 + 1))/(((𝑡^2 )^2 + 2𝑡^2 + 1^2 ) )
𝑑𝑦/𝑑𝑡 = (4𝑡(𝑡^2 + 1))/(𝑡^2 + 1)^2
𝑑𝑦/𝑑𝑡 = 4𝑡/((𝑡^2 + 1) )
Finding 𝒅𝒙/𝒅𝒕
x = tan^(−1)𝑡
𝑑𝑥/𝑑𝑡 = 1/(1 + 𝑡^2 )
Now,
𝑑𝑦/𝑑𝑥 = (𝑑𝑦/𝑑𝑡)/(𝑑𝑥/𝑑𝑡)
𝑑𝑦/𝑑𝑥 = (4𝑡/((𝑡^2 + 1) ))/(1/(1 + 𝑡^2 ))
𝑑𝑦/𝑑𝑥 = 4𝑡/((𝑡^2 + 1) )×((1 + 𝑡^2))/1
𝑑𝑦/𝑑𝑥 = 4t
Now,
(𝑑^2 𝑦)/(𝑑𝑥^2 ) = (𝑑(4𝑡))/𝑑𝑥
= (𝑑(4𝑡))/𝑑𝑥 × 𝑑𝑡/𝑑𝑡
= (𝑑(4𝑡))/𝑑𝑡 × 𝑑𝑡/𝑑𝑥
= 4 × 𝑑𝑡/𝑑𝑥
= 4 × 1/(𝑑𝑥/𝑑𝑡) = 4 (1 + t2)
As 𝑑𝑥/𝑑𝑡 = 1/(1 + 𝑡^2 )
𝑑𝑡/𝑑𝑥 = (1 + t2)

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