Mathematics - Differentiation Question with Solution | TestHub

Let $t=\ln x$. Then $x=e^t$. Using the chain rule, $\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=f^{\prime }(t)\cdot \frac{1}{x}=\frac{f^{\prime }(\ln x)}{x}$. Now, for the second derivative:$$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{f^{\prime }(\ln x)}{x}\right)=\frac{x\cdot f^{\prime \prime }(\ln x)\cdot \frac{1}{x}-f^{\prime }(\ln x)\cdot 1}{x^2}=\frac{f^{\prime \prime }(\ln x)-f^{\prime }(\ln x)}{x^2}$$ Substitute these into the given differential equation:$$x^2\left(\frac{f^{\prime \prime }(\ln x)-f^{\prime }(\ln x)}{x^2}\right)+x\left(\frac{f^{\prime }(\ln x)}{x}\right)+f(\ln x)=0$$ $$f^{\prime \prime }(\ln x)-f^{\prime }(\ln x)+f^{\prime }(\ln x)+f(\ln x)=0$$ This simplifies to $f^{\prime \prime }(\ln x)+f(\ln x)=0$. Let $t=\ln x$. Then $f^{\prime \prime }(t)+f(t)=0$ for all $t$ in the domain. We need to find $f^{\prime \prime }(0)$. From the derived relation, $f^{\prime \prime }(0)+f(0)=0$. Given $f(0)=1$, we have $f^{\prime \prime }(0)+1=0$, which implies $f^{\prime \prime }(0)=-1$.