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\includegraphics[max width=\textwidth, alt={}, center]{a74e4ddf-d254-45f3-bd9a-adf7cd53b3a6-4_611_895_255_625}
The diagram shows the curve \(y = \sqrt { } \left( \frac { 1 - x } { 1 + x } \right)\).
- By first differentiating \(\frac { 1 - x } { 1 + x }\), obtain an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\). Hence show that the gradient of the normal to the curve at the point \(( x , y )\) is \(( 1 + x ) \sqrt { } \left( 1 - x ^ { 2 } \right)\).
- The gradient of the normal to the curve has its maximum value at the point \(P\) shown in the diagram. Find, by differentiation, the \(x\)-coordinate of \(P\).