Question

If p is a polynomial of 3 degree, $y^2 = p(x)$ and $Y = (tan^{-1}x)^2$, then which of the following options is/are equal to

$2\frac{d}{dx}(y^3\frac{d^2y}{dx^2})\times((x^2+1)^2\frac{d^2Y}{dx^2}+2x(1+x^2)\frac{dY}{dx})$?

• A. $2(p'(x) + p'''(x))$
• B. $2p''(x)p'''(x)$
• C. $2 p(x)p'''(x)$
• D. $2 p'(x)p''(x)$

Answer 1

Answer: (C)

$2 p(x)p'''(x)$

Answer 2

Let the given expression be $E$. We need to evaluate $E = 2\frac{d}{dx}(y^3\frac{d^2y}{dx^2})\times((x^2+1)^2\frac{d^2Y}{dx^2}+2x(1+x^2)\frac{dY}{dx})$.

First, consider the term $2\frac{d}{dx}(y^3\frac{d^2y}{dx^2})$. We are given $y^2 = p(x)$. Differentiating with respect to $x$: $2y \frac{dy}{dx} = p'(x)$ Differentiating again with respect to $x$: $2y \frac{d^2y}{dx^2} + 2 (\frac{dy}{dx})^2 = p''(x)$ Differentiating a third time with respect to $x$: $2y \frac{d^3y}{dx^3} + 2 \frac{dy}{dx} \frac{d^2y}{dx^2} + 4 \frac{dy}{dx} \frac{d^2y}{dx^2} = p'''(x)$ $2y \frac{d^3y}{dx^3} + 6 \frac{dy}{dx} \frac{d^2y}{dx^2} = p'''(x)$

Now, evaluate $\frac{d}{dx}(y^3\frac{d^2y}{dx^2})$: $\frac{d}{dx}(y^3\frac{d^2y}{dx^2}) = 3y^2 \frac{dy}{dx} \frac{d^2y}{dx^2} + y^3 \frac{d^3y}{dx^3}$ So, $2\frac{d}{dx}(y^3\frac{d^2y}{dx^2}) = 2(3y^2 \frac{dy}{dx} \frac{d^2y}{dx^2} + y^3 \frac{d^3y}{dx^3}) = 6y^2 \frac{dy}{dx} \frac{d^2y}{dx^2} + 2y^3 \frac{d^3y}{dx^3}$ Factor out $y^2$: $2\frac{d}{dx}(y^3\frac{d^2y}{dx^2}) = y^2 (6 \frac{dy}{dx} \frac{d^2y}{dx^2} + 2y \frac{d^3y}{dx^3})$ From the third derivative equation, $6 \frac{dy}{dx} \frac{d^2y}{dx^2} + 2y \frac{d^3y}{dx^3} = p'''(x)$. Substituting this, we get: $2\frac{d}{dx}(y^3\frac{d^2y}{dx^2}) = y^2 p'''(x)$ Since $y^2 = p(x)$, we have: $2\frac{d}{dx}(y^3\frac{d^2y}{dx^2}) = p(x) p'''(x)$.

Next, consider the term $((x^2+1)^2\frac{d^2Y}{dx^2}+2x(1+x^2)\frac{dY}{dx})$. We are given $Y = (\tan^{-1}x)^2$. $\frac{dY}{dx} = \frac{d}{dx}((\tan^{-1}x)^2) = 2 (\tan^{-1}x) \cdot \frac{1}{1+x^2} = \frac{2 \tan^{-1}x}{1+x^2}$. Let the second term be $S = ((x^2+1)^2\frac{d^2Y}{dx^2}+2x(1+x^2)\frac{dY}{dx})$. We can factor out $(x^2+1)$: $S = (x^2+1) \left[ (x^2+1)\frac{d^2Y}{dx^2} + 2x\frac{dY}{dx} \right]$. The expression in the square bracket is the derivative of the product $(x^2+1)\frac{dY}{dx}$: $\frac{d}{dx}((x^2+1)\frac{dY}{dx}) = (x^2+1)\frac{d^2Y}{dx^2} + 2x\frac{dY}{dx}$. Let's evaluate $(x^2+1)\frac{dY}{dx}$: $(x^2+1)\frac{dY}{dx} = (x^2+1) \left(\frac{2 \tan^{-1}x}{1+x^2}\right) = 2 \tan^{-1}x$. Now, differentiate this with respect to $x$: $\frac{d}{dx}((x^2+1)\frac{dY}{dx}) = \frac{d}{dx}(2 \tan^{-1}x) = \frac{2}{1+x^2}$. So, the expression in the square bracket is $\frac{2}{1+x^2}$. Substituting this back into $S$: $S = (x^2+1) \left[ \frac{2}{1+x^2} \right] = 2$.

Finally, the given expression $E$ is the product of the two parts: $E = \left( 2\frac{d}{dx}(y^3\frac{d^2y}{dx^2}) \right) \times \left( (x^2+1)^2\frac{d^2Y}{dx^2}+2x(1+x^2)\frac{dY}{dx} \right)$ $E = (p(x) p'''(x)) \times (2)$ $E = 2 p(x) p'''(x)$.

Comparing this with the given options: (A) $2(p'(x) + p'''(x))$ (B) $2p''(x)p'''(x)$ (C) $2 p(x)p'''(x)$ (D) $2 p'(x)p''(x)$

The result matches option (C).

The final answer is $\boxed{2 p(x)p'''(x)}$.

The final answer is $\boxed{2 p(x)p'''(x)}$. The calculation for the first part $2\frac{d}{dx}(y^3\frac{d^2y}{dx^2})$ relies on the derivatives of $y$ obtained from $y^2=p(x)$. The calculation for the second part $((x^2+1)^2\frac{d^2Y}{dx^2}+2x(1+x^2)\frac{dY}{dx})$ relies on the derivatives of $Y=(\tan^{-1}x)^2$. The structure of the expression allows for simplification by recognizing it as $(x^2+1) \times \frac{d}{dx}((x^2+1)\frac{dY}{dx})$.