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A tricky example of second derivative

Nalin Pithwa

April 1

Problem : If $ax^{2} + 2hxy + by^{2}=0$ , prove that $\frac{d^{2}y}{dx^{2}}=0$

Solution: Brute force makes it ugly. It won't work. Below is a nice trick.

Given that $ax^{2} + 2hxy + by^{2} = 0$ , differentiating both sides w.r.t. x, we get the following:

$2ax + 2hx \frac{dy}{dx} + 2hy + 2by \frac{dy}{dx} = 0$

Hence, we get : $ax + hx\frac{dy}{dx} + by \frac{dy}{dx} + hy =0$

$(hx + by)\frac{dy}{dx} = -(ax + hy)$

$\frac{dy}{dx} = - \frac{ax+hy}{hx+by}$

So we get $ax^{2} + hxy + hxy + by^{2} = 0$

x(ax + hy) +y(hx + by) = 0

x(ax+by) = -y(hx + by)

$-\frac{ax+by}{hx + by} = \frac{y}{x}$

Hence, we get the following simplification: $\frac{dy}{dx} = \frac{y}{x}$

Differentiating both sides w.r.t. x of the above,

$\frac{d^{2}y}{dx^{2}} = \frac{x\frac{dy}{dx}-y}{x^{2}} = \frac{x \times \frac{y}{x} - y}{x^{2}}$

which in turn equals $\frac{y-y}{x^{2}}=0$ .

Problem 2:

If $y = \sin {(m\arccos{x})}$ prove that $(1-x^{2})\frac{d^{2}}{dy^{2}} - x \frac{dy}{dx} + m^{2}y=0$ .

This is a famous classic question. Found in almost all good IITJEE mathematics books.

Solution:

Given that $y = \sin{(m \arccos{x})}$

So we get $\arcsin{y} = m \times \arccos{x}$

Differentiating both sides w.r.t. x, we get

$\frac{1}{\sqrt{1-y^{2}}} \frac{dy}{dx} = \frac{-m}{\sqrt{1-x^{2}}}$

$\sqrt{1-x^{2}} \frac{dy}{dx} = -m \times \sqrt{1-y^{2}}$

Squaring both sides we get

$(1-x^{2})(\frac{dy}{dx})^{2} = \frac{-m}{\sqrt{1-y^{2}}}$

Differentiating both sides w.r.t. x,

$(1-x^{2})2\frac{dy}{dx} \times \frac{d^{2}y}{dx^{2}} + (-2x) \times (\frac{dy}{dx})^{2} = m^{2}(1-y^{2})$

Dividing both sides by $2 \times \frac{dy}{dx}$ , we get

$(1-x^{2}) \times \frac{d^{2}y}{dx^{2}} -x \frac{dy}{dx} + m^{2}y = 0$

Problem 3:

If $y = (\log ({x + \sqrt{x^{2}+1}}))^{2}$ , prove that $(1+x^{2})\frac{d^{2}y}{dx^{2}} +x\frac{dy}{dx} = 2$

Proof:

Given that $y = (\log {(x + \sqrt{x^{2}+1})})^{2}$ ...call this (I)

Then, $\frac{dy}{dx} = 2 \log {(x+\sqrt{x^{2}+1})} \times (\frac{1}{1+\sqrt{x+\sqrt{x^{2}+1}}} \times (1+ \frac{2x}{2\sqrt{x^{2}+1}}))$

which in turn simplifies to

$2\log {(x + \sqrt{x^{2}+1})} \times (\frac{1}{x+\sqrt{x^{2}+1}} \times (\frac{\sqrt{x^{2}+x}}{\sqrt{x^{2}+1}}))$

which boils down to

$\frac{2\log{(x+\sqrt{x^{2}+1})}}{\sqrt{x^{2}+1}}$

Hence, $\sqrt{x^{2}+1}\frac{dy}{dx} = 2 \log {(x+\sqrt{x^{2}+1})}$

Squaring both sides

$(x^{2}+1)(\frac{dy}{dx})^{2} = 4 \times (\log (x + \sqrt{x^{2}+1}))^{2}$

$(1+x^{2}) \times (\frac{dy}{dx})^{2} = 4 \times (\log (x + \sqrt{x^{2}+1}))^{2}$

$(1+x^{2}) (\frac{dy}{dx})^{2} = 4 \times y$

Differentiate both sides w.r.t. x,

$(1+x^{2}) \times 2 \times \frac{dy}{dx} \times \frac{d^{2}y}{dx^{2}} + 2x \times (\frac{dy}{dx})^{2} = 4 \times \frac{dy}{dx}$

Dividing by $2\frac{dy}{dx}$ , we get

$(1+x^{2}) \times \frac{d^{2}y}{dx^{2}} + x \frac{dy}{dx} = 2$

Hope you enjoyed the above examples just for the sheer joy of playing with math ! IITJEE apart !

Regards,

Nalin Pithwa

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A tricky example of second derivative

A tricky example of second derivative

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