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Y=e^tan(inverse)X,prove that (1+ X²)d2y/dx2 (2x-1) dy/dX=0?
Most Upvoted Answer
Y=e^tan(inverse)X,prove that (1+ X²)d2y/dx2 (2x-1) dy/dX=0?
Actually it should be (1+x^2)^2
and then all gets simple...
just differentiate it once and then twice and put the values in and you'll get the answer...
Community Answer
Y=e^tan(inverse)X,prove that (1+ X²)d2y/dx2 (2x-1) dy/dX=0?


Proof:

Given: Y = e^(tan^(-1)X)

Step 1: Find the first and second derivatives of Y with respect to X.
Y = e^(tan^(-1)X)
Taking natural logarithm on both sides:
ln(Y) = tan^(-1)X
Differentiating both sides with respect to X:
(1/Y) * dY/dX = 1/(1+X^2)
dY/dX = Y/(1+X^2)
Now, differentiate dY/dX with respect to X:
d^2Y/dX^2 = (dY/dX) * (-2X/(1+X^2)^2)
d^2Y/dX^2 = -2X * Y/(1+X^2)^2

Step 2: Substitute the derivatives into the given expression.
(1+X^2) * d^2Y/dX^2 * (2X-1) * dY/dX
= (1+X^2) * (-2X * Y/(1+X^2)^2) * (2X-1) * Y/(1+X^2)
= -2X * Y * (2X-1) * Y/(1+X^2)
= -2X(2X-1) * Y^2/(1+X^2)
= -4X^2Y^2 + 2XY^2

Step 3: Simplify the expression.
Since Y = e^(tan^(-1)X), we can substitute this into the expression:
-4X^2(e^(tan^(-1)X))^2 + 2Xe^(tan^(-1)X)
= -4X^2(e^(2(tan^(-1)X))) + 2Xe^(tan^(-1)X)
= -4X^2(e^(tan^(-1)(X))^2) + 2Xe^(tan^(-1)X)
= 0

Therefore, (1+X^2)d^2Y/dX^2 * (2X-1) * dY/dX = 0.
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Y=e^tan(inverse)X,prove that (1+ X²)d2y/dx2 (2x-1) dy/dX=0?
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