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How do you find the second derivative of #2x^2-y^2=1#?

Answer 1

#(d^2y)/dx^2=-2/y^3#

We have #2x^2-y^2=1#.........equation 1 Differentiating with respect to #x# we get #4x-2y (dy)/(dx) =0# #=>(dy)/(dx)=2x/y# Again differentiating with respect to #x# we get #(d^2y)/dx^2=2 ((y-x (dy)/(dx)))/y^2# Now putting the value of #(dy)/(dx)# we get #(d^2y)/dx^2=2( (y^2-2x^2))/y^3# #:.(d^2y)/dx^2=-2/y^3# (using equation 1)

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Answer 2

Axel Alvarez

To find the second derivative of (2x^2 - y^2 = 1), you first need to implicitly differentiate the equation twice with respect to (x).

Find the first derivative: [4x - 2yy' = 0] [y' = \frac{4x}{2y}]
Find the second derivative using the quotient rule: [y'' = \frac{d}{dx} \left(\frac{4x}{2y}\right)] [= \frac{(2y)(4) - (4x)(2y')}{(2y)^2}] [= \frac{8y - 8xy'}{4y^2}] [= \frac{8y - 8x\left(\frac{4x}{2y}\right)}{4y^2}] [= \frac{8y - \frac{32x^2}{y}}{4y^2}]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.