Question

If $\log y=\log(\sin x)-x^2$, then \[ \frac{d^2y}{dx^2}+4x\frac{dy}{dx}+4x^2y= \]

• A. $-2y$
• B. $3y$
• C. $0$
• D. $-3y$

Hint:

Whenever a function contains $e^{f(x)}$, try expressing derivatives back in terms of the original function $y$. This significantly reduces lengthy differentiation steps and simplifies higher-order derivatives.

Answer 1

The Correct Option is D

Concept: This problem is based on logarithmic differentiation and higher order differentiation. The given equation is first simplified into an explicit form of $y$, after which successive differentiation is performed carefully. The important ideas involved are:
• Using logarithmic properties to simplify the expression.
• Converting logarithmic form into exponential form.
• Applying product rule and chain rule repeatedly.
• Expressing derivatives in terms of the original function $y$ to simplify calculations.

Step 1: Convert the logarithmic equation into explicit form.
Given, \[ \log y=\log(\sin x)-x^2 \] Using the logarithmic identity, \[ \log a-\log b=\log\left(\frac{a}{b}\right) \] we rewrite: \[ \log y=\log\left(\sin x\right)-\log(e^{x^2}) \] Thus, \[ \log y=\log\left(\frac{\sin x}{e^{x^2}}\right) \] Taking antilogarithm on both sides: \[ y=\frac{\sin x}{e^{x^2}} \] Hence, \[ y=e^{-x^2}\sin x \] This explicit form will now be differentiated.

Step 2: Find the first derivative $\dfrac{dy}{dx}$.
We differentiate \[ y=e^{-x^2}\sin x \] using the product rule: \[ \frac{d}{dx}(uv)=u\frac{dv}{dx}+v\frac{du}{dx} \] Let \[ u=e^{-x^2}, \qquad v=\sin x \] Then, \[ \frac{du}{dx}=e^{-x^2}(-2x) \] and \[ \frac{dv}{dx}=\cos x \] Therefore, \[ \frac{dy}{dx}=e^{-x^2}\cos x+\sin x\left(-2xe^{-x^2}\right) \] \[ \frac{dy}{dx}=e^{-x^2}\cos x-2xe^{-x^2}\sin x \] Since \[ e^{-x^2}\sin x=y \] we get \[ \frac{dy}{dx}=e^{-x^2}\cos x-2xy \] Hence, \[ e^{-x^2}\cos x=\frac{dy}{dx}+2xy \] This relation will help simplify the second derivative.

Step 3: Find the second derivative $\dfrac{d^2y}{dx^2}$.
Differentiate \[ \frac{dy}{dx}=e^{-x^2}\cos x-2xy \] Differentiating term by term: \[ \frac{d^2y}{dx^2}=\frac{d}{dx}\left(e^{-x^2}\cos x\right)-\frac{d}{dx}(2xy) \] Using product rule on the first term: \[ \frac{d}{dx}\left(e^{-x^2}\cos x\right) =e^{-x^2}(-\sin x)+\cos x\left(-2xe^{-x^2}\right) \] Thus, \[ =e^{-x^2}(-\sin x)-2xe^{-x^2}\cos x \] Since \[ e^{-x^2}\sin x=y \] we get \[ =-y-2xe^{-x^2}\cos x \] Now differentiate the second term: \[ \frac{d}{dx}(2xy)=2x\frac{dy}{dx}+2y \] Therefore, \[ \frac{d^2y}{dx^2} =-y-2xe^{-x^2}\cos x-2x\frac{dy}{dx}-2y \] Combining like terms: \[ \frac{d^2y}{dx^2} =-3y-2xe^{-x^2}\cos x-2x\frac{dy}{dx} \] Now substitute \[ e^{-x^2}\cos x=\frac{dy}{dx}+2xy \] So, \[ \frac{d^2y}{dx^2} =-3y-2x\left(\frac{dy}{dx}+2xy\right)-2x\frac{dy}{dx} \] Expanding: \[ \frac{d^2y}{dx^2} =-3y-2x\frac{dy}{dx}-4x^2y-2x\frac{dy}{dx} \] \[ \frac{d^2y}{dx^2} =-3y-4x\frac{dy}{dx}-4x^2y \]

Step 4: Evaluate the required expression.
Rearranging: \[ \frac{d^2y}{dx^2}+4x\frac{dy}{dx}+4x^2y=-3y \] Hence, \[ \boxed{-3y} \]