Q.If $y = \sin^{-1} x$, show that $(1 - x^2)\frac{d^2y}{dx^2} - x\frac{dy}{dx} = 0$.

Concept and Intuition

The expression inside the derivative looks like a ratio of two trigonometric terms that are actually conjugates in disguise. Instead of diving straight into the quotient rule (which would be messy), we can simplify first by multiplying numerator and denominator by something clever, or by noticing a known identity:

This is because $(\sec x + \tan x)(\sec x - \tan x) = \sec^2 x - \tan^2 x = 1$.

So the fraction is actually

That’s much easier to differentiate.

Step-by-step

1. Simplify the fraction Using the identity $\sec^2 x - \tan^2 x = 1$, we have

because multiplying numerator and denominator by $(\sec x + \tan x)$ gives

2. Differentiate Let $f(x) = (\sec x + \tan x)^2$. Then

Factor $\sec x$ from the derivative of the inside:

So

3. Evaluate at $x = \pi/4$ At $\pi/4$, $\sec x = \sqrt{2}$ and $\tan x = 1$, so

Hence

Compute $(\sqrt{2} + 1)^2 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2}$.

So

$k = 2\sqrt{2} (3 + 2\sqrt{2}) = 6\sqrt{2} + 8$ …

Given $y=f(\cosh x)$ with $f'(x)=\log\!\left(x+\sqrt{x^2-1}\right)=\cosh^{-1}x$.

First derivative (chain rule). With $g(x)=\cosh x$,

Since $f'(t)=\cosh^{-1}t$ and $\cosh^{-1}(\cosh x)=x$ (for $x\ge 0$),

Second derivative (product rule). …

The key insight: an infinite nested radical of the form

can be rewritten as $y = \sqrt{u + y}$, because the pattern repeats indefinitely. This turns an intimidating infinite expression into a simple algebraic equation, which we can then differentiate.

Here, $u = \log(x^2+1)$. So we have:

1. Square both sides to remove the square root:

2. Rearrange to isolate the logarithmic term:

3. Differentiate implicitly with respect to $x$. Remember that $y$ is a function of $x$:

This gives:

4. Factor out $\frac{dy}{dx}$ on the left:

5. Solve for $\frac{dy}{dx}$:

--- …

The key idea is that the derivative of $\sec^{-1} x$ has a standard form involving an absolute value, because the range of $\sec^{-1} x$ is chosen to make the function one-to-one. When we differentiate a second time, we must treat $|x|$ correctly — it is not simply $x$, and its derivative introduces a sign factor. The algebra then simplifies to one of the given forms.

1. Recall the first derivative. For $y = \sec^{-1} x$, the derivative is

The absolute value appears because the inverse secant is defined with range $[0, \pi/2) \cup (\pi/2, \pi]$, and the slope of the tangent must be positive for $x>1$ and negative for $x<-1$.

2. Rewrite the absolute value for differentiation. A useful trick: $|x| = \sqrt{x^2}$. Then

This avoids piecewise cases during differentiation, but we must remember that $\sqrt{x^2} = |x|$, not $x$.

3. Differentiate again using the chain rule. Let $u = x^2(x^2 - 1) = x^4 - x^2$. Then

So

Compute $\frac{du}{dx} = 4x^3 - 2x = 2x(2x^2 - 1)$.

4. Substitute back.

The factor $-\frac{1}{2} \cdot 2 = -1$, so

5. Simplify the denominator.

Since $|x|^3 = |x| \cdot x^2$ (because $|x|^3 = |x| \cdot x^2$), we have

6. Handle the sign from $x/|x|$. Notice $x/|x| = \operatorname{sgn}(x)$, but here we have $x/|x|^3 = \frac{x}{|x| \cdot x^2} = \frac{\operatorname{sgn}(x)}{x^2}$? Wait carefully:

But we want the expression in the form given in the options. Multiply numerator and denominator:

Actually, from step 5:

Write $|x|^3 = |x| \cdot x^2$ (since $|x|^3 = |x| \cdot x^2$ for all real $x$). Then

Now $x/|x| = \operatorname{sgn}(x)$, so

But the options have $x|x|$ in the denominator, not $x^2$ times a sign. Multiply numerator and denominator by $|x|$:

Because $x^2 \cdot |x| = |x|^3$? Wait: $x|x|$ in denominator times $|x|$ gives $x|x|^2 = x \cdot x^2 = x^3$? That’s messy. Let’s check directly: …

The key insight is that an infinitely nested expression like this is self‑referential: the whole infinite chain equals $y$, and the part after the first square root is the same infinite chain again. That lets us replace the infinite tail with $y$ itself, turning a scary infinite radical into a simple algebraic equation.

1. Set up the self‑referential equation Since the pattern repeats infinitely, we have

Therefore

2. Square both sides Squaring removes the outer square root:

Rearranging gives

3. Differentiate implicitly Treat $y$ as a function of $x$. Differentiate term‑by‑term:

Using the chain rule:

4. Solve for $\frac{dy}{dx}$ Factor out $\frac{dy}{dx}$: $$$$

Differentiating $y = \log\!\left(x - \sqrt{x^2-1}\right)$.

Let $u = x - \sqrt{x^2-1}$, so $y = \log u$. Then

Hence

Differentiate again:

Combining the derivative terms.

$(x^2-1)y'' + xy' = \frac{x}{\sqrt{x^2-1}} - \frac{x}{\sqrt{x^2-1}} = 0.$ …

Concept and intuition

Order is simply the highest derivative present — here we see both $\frac{dy}{dx}$ (first derivative) and $\frac{d^2y}{dx^2}$ (second derivative), so the order is clearly 2.

Degree is trickier: it is the exponent of the highest-order derivative after the equation is made polynomial in derivatives (no radicals, no fractional powers). The given equation has a square root containing $\frac{d^2y}{dx^2}$, so we must eliminate that radical before reading off the degree. Squaring will introduce a term like $\left(\frac{d^2y}{dx^2}\right)^2$, making the degree 2 — but we must check that no higher power appears.

Step-by-step solution

1. Identify the order

   The highest derivative in the equation is $\frac{d^2y}{dx^2}$. Therefore the order is $2$.

2. Prepare to find the degree

   The equation is

The term $\left( \frac{d^2y}{dx^2} + 2 \right)^{1/2}$ is a radical involving the second derivative. To find the degree, we must rewrite the equation so that all derivatives appear with integer exponents.

3. Isolate the radical Move the non‑radical terms to the left:

4. Square both sides Squaring eliminates the square root: $$$$

The problem asks for the derivative of a product of two functions:

$u(x) = x^{5/2} - x^{3/2} + 1$ and $v(x) = x^2 - 3x + 5$.

The product rule is the natural choice: $\frac{d}{dx}[u \cdot v] = u'v + uv'$.

We could expand the product first and then differentiate, but that would involve multiplying two polynomials with fractional exponents — messy. The product rule keeps the algebra cleaner because we differentiate each factor separately (both are straightforward power-rule jobs) and then combine like terms.

Let’s do it step by step.

1. Differentiate $u(x)$

   $u(x) = x^{5/2} - x^{3/2} + 1$

   Using $\frac{d}{dx} x^n = n x^{n-1}$:

   $u'(x) = \frac{5}{2} x^{3/2} - \frac{3}{2} x^{1/2} + 0 = \frac{5}{2} x^{3/2} - \frac{3}{2} x^{1/2}$.

2. Differentiate $v(x)$

   $v(x) = x^2 - 3x + 5$

   $v'(x) = 2x - 3$.

3. Apply the product rule

   $\frac{d}{dx}[u v] = u' v + u v'$

4. Expand the first product $u'v$

So $u'v = \frac{5}{2} x^{7/2} + \left(-\frac{15}{2} - \frac{3}{2}\right) x^{5/2} + \left(\frac{25}{2} + \frac{9}{2}\right) x^{3/2} - \frac{15}{2} x^{1/2}$

5. Expand the second product $uv'$

$1 \cdot 2x = 2x, \quad 1 \cdot (-3) = -3.$ …

We are given $f(0)=0$, $f'(0)=20$, and $f$ is differentiable. The expression $A(x)$ involves $\csc 4x$, which blows up as $x\to 0$, so the limit is not obvious by direct substitution. The key is to rewrite the trigonometric terms near $x=0$ and use the derivative condition.

1. Rewrite $A(x)$ in a form that exposes the limit. Since $\csc 4x = \frac{1}{\sin 4x}$, we have

As $x\to 0$, $\sin 4x \sim 4x$, and $f(x)\sim f'(0)x = 20x$, so the first term behaves like $\frac{2\cdot 20x}{4x}=10$, which is finite. The other terms are finite too, so the limit exists.

2. Combine terms to apply L'Hôpital's rule cleanly. Write

As $x\to 0$, $\sin 4x \to 0$ and $f(x)\to 0$, so the first term is $0/0$. The rest are well-behaved. To handle the $0/0$ part, we isolate it: