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Given that $y=\arctan(x^2)$ find $\ \dfrac{d^2y}{dx^2}$.

I got

$$\frac{dy}{dx}=\frac{2x}{1+x^4}.$$

Using low d high minus high d low over low squared, I got

$$\frac{d^2y}{dx^2}=\frac{(1+x)^4 \cdot 2 - 2x \cdot 4(1+x)^3}{(1+x^4)^2}.$$

I tried to simplify this but didn't get the answer which is

$$\frac{d^2y}{dx^2}=\frac{2(1-3x^4)}{(1+x^4)^2}.$$

Where am I going wrong?

Tunk-Fey
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Jim
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  • You treated the denominator as though it were $(1 + x)^4$, whereas it is given as $(1 + x^4)$ when you took the second derivative, though you got it right when you squared the denominator. – amWhy May 16 '14 at 19:01

3 Answers3

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You did the low d high minus high d low wrong: The numerator should be:

$$(1+x^4)2 - (2x)(4x^3)$$

Paul Hurst
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Once you get to the second step, you can indeed use the chain rule. Perhaps if you write out what you intend to find, rather than simply finding it, it'll become easier to see:

$$\left(\frac{2x}{1+x^4} \right)' = \frac{(2x)'\left(1+x^4\right) + (2x)\left( 1+x^4\right)'}{\left(1 + x^4 \right)^2}$$

dmk
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Alternatively,

$ \large \tan (y) = x^2 \Rightarrow \sec^2 (y) \cdot \frac { \mathrm{d}y}{\mathrm{d}x} = 2x $

$ \large \Rightarrow \frac { \mathrm{d}y}{\mathrm{d}x} = \frac {2x}{\sec^2 (y)} = \frac {2x}{\tan^2 (y) + 1} = \frac {2x}{x^4 + 1} $

$ \large \Rightarrow \frac { \mathrm{d^2}y}{\mathrm{d}x^2} = \frac {2(x^4+ 1) - 2x(4x^3)}{(x^4+1)^2} = \frac {-6x^4 + 2}{(x^4 + 1)^2} $

GohP.iHan
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