Find the derivative of y - t2 - 1t2 - 1-

We apply the quotient rule with u = t^2 1 and v = t^2 + 1 . dydt = ddt (uv ) = v(dudt) u(dvdt)v^2 = (t^2 + 1) . 2t (t^2 1) . 2t(t^2 ...

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Basic Differentiation Rule

Find the deri...

Question

Find the derivative of $y = \frac{t^{2} - 1}{t^{2} + 1}$ .

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Solution

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y = \frac{(t^{2} - 2)}{(t^{2} + 2)}

\frac{d y}{d x}

x = \frac{1}{1 - t^{2}}

y = 1 + t^{2}

x = \sqrt{\frac{1 - t^{2}}{1 + t^{2}}}

y = \frac{\sqrt{1 + t^{2}} - \sqrt{1 - t^{2}}}{\sqrt{1 + t^{2}} + \sqrt{1 - t^{2}}},

\frac{d^{2} y}{d x^{2}}

t = 0

\frac{d y}{d x}

x = \frac{1 - t^{2}}{1 + t^{2}} and y = \frac{2 t}{1 + t^{2}}

\frac{d y}{d x}

x = \frac{2 t}{1 + t^{2}} and y = \frac{1 - t^{2}}{1 + t^{2}}

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