Mathematics - Differentiation Question with Solution | TestHub

Let $f(x)=\frac{x}{1+x^2}$.$f^{\prime }(x)=\frac{(1+x^2)(1)-x(2x)}{(1+x^2{)}^2}=\frac{1-x^2}{(1+x^2{)}^2}$.$f^{\prime \prime }(x)=\frac{(1+x^2{)}^2(-2x)-(1-x^2)\cdot 2(1+x^2)(2x)}{(1+x^2{)}^4}=\frac{(1+x^2)(-2x)-4x(1-x^2)}{(1+x^2{)}^3}=\frac{2x^3-6x}{(1+x^2{)}^3}$. At $x=0$: $f(0)=0$, $f^{\prime }(0)=1$, $f^{\prime \prime }(0)=0$. Let $y=f(g(x))$ where $g(x)=f(f(x))$.$\frac{dy}{dx}=f^{\prime }(g(x))g^{\prime }(x)$.$\frac{d^{2}y}{d{x}^2}=f^{\prime \prime }(g(x))(g^{\prime }(x){)}^2+f^{\prime }(g(x))g^{\prime \prime }(x)$. At $x=0$: $g(0)=f(f(0))=f(0)=0$.$g^{\prime }(x)=f^{\prime }(f(x))f^{\prime }(x)⟹g^{\prime }(0)=f^{\prime }(f(0))f^{\prime }(0)=f^{\prime }(0)f^{\prime }(0)=(1)(1)=1$.$g^{\prime \prime }(x)=f^{\prime \prime }(f(x))(f^{\prime }(x){)}^2+f^{\prime }(f(x))f^{\prime \prime }(x)⟹g^{\prime \prime }(0)=f^{\prime \prime }(f(0))(f^{\prime }(0){)}^2+f^{\prime }(f(0))f^{\prime \prime }(0)=f^{\prime \prime }(0)(1{)}^2+f^{\prime }(0)(0)=0(1)+1(0)=0$. Substitute these values into $\frac{d^{2}y}{d{x}^2}$ at $x=0$:$\frac{d^{2}y}{d{x}^2}{|}_{x=0}=f^{\prime \prime }(g(0))(g^{\prime }(0){)}^2+f^{\prime }(g(0))g^{\prime \prime }(0)=f^{\prime \prime }(0)(1{)}^2+f^{\prime }(0)(0)=0(1)+1(0)=0$.