IB Maths AA HL 5.15 Further derivatives and integrals Question Bank

attempting to use the quotient rule to find $\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{x^{2}-1}{x^{2}+1}\right)$$\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{x^{2}-1}{x^{2}+1}\right)=\frac{2 x\left(x^{2}+1\right)-2 x\left(x^{2}-1\right)}{\left(x^{2}+1\right)^{2}}\left(=\frac{4 x}{\left(x^{2}+1\right)^{2}}\right)$
attempting to use the chain rule to find $\frac{\mathrm{d}}{\mathrm{dx}}\left(\arcsin \left(\frac{x^{2}-1}{x^{2}+1}\right)\right)$
let $u=\frac{x^{2}-1}{x^{2}+1}$ and so $y=\arcsin u$ and $\frac{\mathrm{d} y}{\mathrm{~d} u}=\frac{1}{\sqrt{1-u^{2}}}$$f^{\prime}(x)=\frac{1}{\sqrt{1-\left(\frac{x^{2}-1}{x^{2}+1}\right)^{2}}} \times \frac{4 x}{\left(x^{2}+1\right)^{2}}$$=\frac{4 x}{\sqrt{\left(x^{2}+1\right)^{2}-\left(x^{2}-1\right)^{2}}} \times \frac{1}{\left(x^{2}+1\right)}$$=\frac{4 x}{\sqrt{4 x^{2}}} \times \frac{1}{\left(x^{2}+1\right)}$$=\frac{2 x}{\sqrt{x^{2}}\left(x^{2}+1\right)}$