Mathematics - Differentiation Question with Solution | TestHub

First, differentiate the implicit equation $y^3+xy+x^3=3$ with respect to $x$:$$3y^2\frac{dy}{dx}+(1\cdot y+x\frac{dy}{dx})+3x^2=0$$ $$3y^2\frac{dy}{dx}+y+x\frac{dy}{dx}+3x^2=0$$ Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(3y^2+x)=-(y+3x^2)$. So, $f^{\prime }(x)=\frac{-(f(x)+3x^2)}{3(f(x){)}^2+x}$. Next, for $g(x)=f(x^2+x)$, find $g^{\prime }(x)$ using the chain rule:$$g^{\prime }(x)=f^{\prime }(x^2+x)\cdot \frac{d}{dx}(x^2+x)=f^{\prime }(x^2+x)\cdot (2x+1)$$ We need to evaluate $g^{\prime }(0)$: $g^{\prime }(0)=f^{\prime }(0^2+0)\cdot (2(0)+1)=f^{\prime }(0)$. To find $f^{\prime }(0)$, we first need $f(0)$. Substitute $x=0$ into the original implicit equation:$f(0{)}^3+(0)f(0)+(0{)}^3=3⟹f(0{)}^3=3⟹f(0)=\sqrt[3]{3}$. Now substitute $x=0$ and $f(0)=\sqrt[3]{3}$ into the expression for $f^{\prime }(x)$:$$f^{\prime }(0)=\frac{-(f(0)+3(0{)}^2)}{3(f(0){)}^2+0}=\frac{-f(0)}{3(f(0){)}^2}=\frac{-1}{3f(0)}$$ Substituting $f(0)=\sqrt[3]{3}$, we get $f^{\prime }(0)=\frac{-1}{3\sqrt[3]{3}}$. Therefore, $g^{\prime }(0)=\frac{-1}{3\sqrt[3]{3}}$.