Q.Find $\frac{d^2y}{dx^2}$, if $y = x^3 + \tan x$.

We start with the function $f(x) = \sin 3x \cos 5x$. Differentiating a product of trig functions directly would involve the product rule twice, which is messy. Instead, we use a trigonometric identity to turn the product into a sum — this makes differentiation straightforward.

Concept & Intuition:

The product-to-sum identity for sine and cosine is:

This converts a product into a sum of two sine functions, each of which is easy to differentiate. After that, we just take the second derivative term by term.

Step-by-step solution:

1. Rewrite the product as a sum. Let $A = 3x$ and $B = 5x$. Then:

Since $\sin(-\theta) = -\sin\theta$, we have:

2. Find the first derivative. Differentiate term by term:

3. Find the second derivative. Differentiate $f'(x)$:

Rearranging:

4. Match with the options. This expression is exactly option (D).

We start with

Exponentiating both sides (using base $e$) gives

So $y = e^{-x^2} \sin x$. This is a product of an exponential decay and a sine wave — a classic damped oscillation.

Now we need $\frac{d^2 y}{d x^2} + 4x \frac{dy}{dx} + 4x^2 y$.

1. First derivative Using the product rule:

2. Second derivative Differentiate $\frac{dy}{dx}$ again, again using product rule on $e^{-x^2}$ times $(\cos x - 2x \sin x)$:

Compute the derivative inside:

So together:

Thus

Factor $e^{-x^2}$:

Simplify:

3. Now form the expression We have:

Substitute each term:

• $\frac{d^2 y}{dx^2} = e^{-x^2}(-3\sin x - 4x \cos x + 4x^2 \sin x)$
• $4x \frac{dy}{dx} = 4x \cdot e^{-x^2}(\cos x - 2x \sin x) = e^{-x^2}(4x \cos x - 8x^2 \sin x)$
• $4x^2 y = 4x^2 \cdot e^{-x^2} \sin x = e^{-x^2}(4x^2 \sin x)$

Add them:

4. Simplify inside the brackets

   • $\sin x$ terms: $-3\sin x + 4x^2 \sin x - 8x^2 \sin x + 4x^2 \sin x = -3\sin x + (4 - 8 + 4)x^2 \sin x = -3\sin x + 0$.
   • $\cos x$ terms: $-4x \cos x + 4x \cos x = 0$.

   So the whole expression becomes:

Thus the expression simplifies to $-3y$.

Let $y=(\sin^{-1}x)^2+(\cos^{-1}x)^2$. Then

Multiply through by $\sqrt{1-x^2}$:

Differentiate both sides. The left side gives $-\dfrac{x}{\sqrt{1-x^2}}\,y'+\sqrt{1-x^2}\,y''$; the right side gives $2\left(\dfrac{1}{\sqrt{1-x^2}}+\dfrac{1}{\sqrt{1-x^2}}\right)=\dfrac{4}{\sqrt{1-x^2}}$. So

Multiplying by $\sqrt{1-x^2}$:

Concept & Intuition

When a function is given as $y$ in terms of $x$, but we need derivatives of $x$ with respect to $y$, we can’t just “flip” the derivative naively. Instead, we use the fact that

provided $\frac{dy}{dx} \neq 0$. For the second derivative, we differentiate $\frac{dx}{dy}$ with respect to $y$, which requires the chain rule because $\frac{dx}{dy}$ is expressed in terms of $x$, and $x$ itself depends on $y$. This avoids solving for $x$ explicitly (which is impossible here anyway).

Step-by-step solution

1. Find $\frac{dy}{dx}$ Given $y = x + e^x$, differentiate with respect to $x$:

2. Find $\frac{dx}{dy}$ Using the reciprocal relation:

3. Set up for $\frac{d^2 x}{dy^2}$ The second derivative is the derivative of $\frac{dx}{dy}$ with respect to $y$:

Since the expression is in terms of $x$, we use the chain rule:

4. Differentiate with respect to $x$ Let $u = 1 + e^x$. Then $\frac{1}{u}$ differentiates to $-\frac{1}{u^2} \cdot \frac{du}{dx}$:

5. Multiply by $\frac{dx}{dy}$ From step 2, $\frac{dx}{dy} = \frac{1}{1 + e^x}$. So:

Concept & Intuition

When a function involves a logarithm of a quotient or power, it’s almost always easier to expand it using properties of logs before differentiating. Here, $y = \log\left(\frac{x^2}{e^2}\right)$ can be rewritten as $\log(x^2) - \log(e^2) = 2\log x - 2$. That turns a messy quotient into a simple difference, making differentiation straightforward. The first derivative will be a simple rational function, and the second derivative follows directly.

Step-by-step solution

1. Simplify the expression Use the logarithm rules:

(Recall $\log(e^2)=2$.) This is much cleaner.

2. First derivative Differentiate term by term:

3. Second derivative Differentiate $\frac{2}{x}$ (which is $2x^{-1}$):

4. Match with options The result $-\frac{2}{x^2}$ corresponds exactly to option (A).

(Note: the first term of the printed stem is a corrupted glyph; the relation is the standard implicit curve $e^{y} + xy = e$, which is the only reading consistent with the four printed options.)

Step 1 — Find $y$ at $x = 0$.

Put $x = 0$ in $e^{y} + xy = e$:

So the point of interest is $(0,\,1)$.

Step 2 — First derivative (implicit differentiation).

Differentiate both sides w.r.t. $x$, using the chain rule on $e^{y}$ and the product rule on $xy$:

At $(0,1)$: $e^{1}\,y' + 1 + 0 = 0$, so

Step 3 — Second derivative.

Differentiate the equation of Step 2 again w.r.t. $x$:

(The first two terms come from differentiating $e^{y}y'$ as a product; the last two from differentiating $y + xy'$.)

Substitute $x = 0$, $y = 1$, $y' = -\tfrac{1}{e}$:

Step 4 — Assemble the ordered pair.

ANSWER: C

Step 1 — Why logarithmic differentiation.

$y$ is a product of three powers. Using the product rule directly on three factors is messy and error-prone. Taking $\log$ converts products into sums and powers into multipliers, which is far cleaner:

Step 2 — Differentiate implicitly.

Using $\dfrac{d}{dx}\log(u) = \dfrac{u'}{u}$ on both sides:

Step 3 — Evaluate $y$ at $x = 4$.

Step 4 — Evaluate the bracket at $x = 4$.

Common denominator 6:

Step 5 — Multiply.

(Note $x=4$ is safely away from $x=1,2,3$, so no denominator blows up and the method is valid.)

ANSWER: D

[!TLDR]

Computing both second derivatives and multiplying gives $-2/y=-2e^{-2x}$, i.e. option (C).

Concept

For a function and its inverse you differentiate each in its own variable; a second derivative with respect to the other variable is found from $x=x(y)$ (CBSE/NCERT Class 12 Continuity and Differentiability).

Solution

With $y=e^{2x}$:

Inverting, $x=\tfrac12\ln y$, so

Multiplying the two second derivatives:

Note that $-2/y$ and $-2e^{-2x}$ are the same value since $y=e^{2x}$; the fully evaluated closed form is $-2e^{-2x}$.

[!ANSWER]

(C) $-2e^{-2x}$ (equivalently $-2/y$)