Mathematics - Differentiation Question with Solution | TestHub

The given differential equation is $f(x)+x{f}^{\prime }(x)=\frac{1}{1+x^2}$. The left side is the derivative of the product $xf(x)$. So, $\frac{d}{dx}(xf(x))=\frac{1}{1+x^2}$. Integrating both sides gives $xf(x)={\tan }^{-1}(x)+C$. Given $f(1)=0$, substitute $x=1$: $1\cdot f(1)={\tan }^{-1}(1)+C\Rightarrow 0=\frac{\pi }{4}+C\Rightarrow C=-\frac{\pi }{4}$. Thus, $xf(x)={\tan }^{-1}(x)-\frac{\pi }{4}$.

We need to find $g^{\prime }(0)$. By the inverse function theorem, $g^{\prime }(y_0)=\frac{1}{f^{\prime }(x_0)}$ where $y_0=f(x_0)$. Here, $y_0=0$. We find $x_0$ such that $f(x_0)=0$. From $xf(x)={\tan }^{-1}(x)-\frac{\pi }{4}$, if $f(x_0)=0$, then $x_0\cdot 0={\tan }^{-1}(x_0)-\frac{\pi }{4}$. This means ${\tan }^{-1}(x_0)=\frac{\pi }{4}$, so $x_0=1$. This matches $f(1)=0$.

Now we find $f^{\prime }(1)$ from the given differential equation: $f(x)+x{f}^{\prime }(x)=\frac{1}{1+x^2}$. Substitute $x=1$: $f(1)+1\cdot f^{\prime }(1)=\frac{1}{1+1^2}$. Since $f(1)=0$, we have $0+f^{\prime }(1)=\frac{1}{2}$, so $f^{\prime }(1)=\frac{1}{2}$. Therefore, $g^{\prime }(0)=\frac{1}{f^{\prime }(1)}=\frac{1}{1/2}=2$.