If $x=a\!\left(\dfrac{1+t^{2}}{1-t^{2}}\right)$ and $y=\dfrac{2t}{1-t^{2}}$, find $\dfrac{dy}{dx}$.

CBSE Class 12 · 2005 · DifferentiationLong Answer

x = a (\frac{1 + t ^{2}}{1 - t ^{2}})

and

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

, find

\frac{d y}{d x}

Answer: $\frac{1 + t ^{2}}{2 a t}$

Differentiate each w.r.t.

t

using the quotient rule.

\frac{d x}{d t} = a \cdot \frac{( 2 t ) ( 1 - t ^{2} ) - ( 1 + t ^{2} ) ( - 2 t )}{( 1 - t ^{2} ) ^{2}} = \frac{4 a t}{( 1 - t ^{2} ) ^{2}}

\frac{d y}{d t} = \frac{2 ( 1 - t ^{2} ) - 2 t ( - 2 t )}{( 1 - t ^{2} ) ^{2}} = \frac{2 ( 1 + t ^{2} )}{( 1 - t ^{2} ) ^{2}}

Divide:

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} = \frac{2 ( 1 + t ^{2} )}{4 a t}

\frac{d y}{d x} = \frac{1 + t ^{2}}{2 a t}

CBSE Class 12 Board · 2005 · Subjective