Question

find dx/dt when y=squareroot(9-2x^2) given x=2 y=1

Answer 1

the derivative of
\[\sqrt{9-2x^2}\] is \[\frac{1}{2\sqrt{9-2x^2}}\times -4x\] by the chain rule
bit of algebra gets
\[-\frac{2x}{\sqrt{9-2x^2}}\]
now replace \(x\) by \(2\)