Calculus, EXT1 EQ-Bank 16 Find the equation of the tangent to the curve \(y=\tan ^{-1}\left(x^2\right)\) at the point on the curve where \(x=1\). (3 marks) --- 8 WORK AREA LINES (style=lined) --- Show Answers Only \(y=x-1+\dfrac{\pi}{4}\) Show Worked Solution \(y=\tan ^{-1}\left(x^2\right) \ \Rightarrow \ \dfrac{dy}{dx}=\dfrac{1}{1+\left(x^2\right)^2} \times 2 x=\dfrac{2 x}{1+x^4}\) \(\text{At} \ \ x=1:\) \(y=\tan ^{-1}\left(1^2\right)=\dfrac{\pi}{4}\) \(\dfrac{dy}{dx}=\dfrac{2}{1+1^4}=1\) \(\text{Find equation of tangent,} \ \ m=1 \ \ \text {through}\ \left(1, \dfrac{\pi}{4}\right):\) \(y-\dfrac{\pi}{4}\) \(=1(x-1)\) \(y\) \(=x-1+\dfrac{\pi}{4}\)