"Let y = y(x) be a function of x satisfying $y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}}$ where k is a constant and $y\left( {{1 \over 2}} \right) = - {1 \over 4}$. Then ${{dy} \over {dx}}$ at x = ${1 \over 2}$, is equal to :"

Question

Accepted Answer

"$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}}$ ....(1)

On differentiating both side of eq. (1) w.r.t. x we get,

${{dy} \over {dx}}\sqrt {1 - {x^2}} - y{{2x} \over {2\sqrt {1 - {x^2}} }}$

= 0 - $\sqrt {1 - {y^2}} + {{xy} \over {\sqrt {1 - {y^2}} }}{{dy} \over {dx}}$

Put x = ${1 \over 2}$ and y = $- {1 \over 4}$, we get

$${{dy} \over {dx}}{{\sqrt 3 } \over 2} - \left( { - {1 \over 4}} \right){{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}$$

= $$ - {{\sqrt {15} } \over 4} + {{ - {1 \over 8}} \over {{{\sqrt {15} } \over 4}}}.{{dy} \over {dx}}$$

$\therefore$ ${{dy} \over {dx}} = - {{\sqrt 5 } \over 2}$"

Let y = y(x) be a function of x satisfying

$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}}$ where k is a constant and

$y\left( {{1 \over 2}} \right) = - {1 \over 4}$. Then ${{dy} \over {dx}}$ at x = ${1 \over 2}$, is equal to :

Solution

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Let y = y(x) be a function of x satisfying $y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}}$ where k is a constant and $y\left( {{1 \over 2}} \right) = - {1 \over 4}$. Then ${{dy} \over {dx}}$ at x = ${1 \over 2}$, is equal to :

Solution

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More Differentiation questions

Drill 25 more like these. Every day. Free.

Let y = y(x) be a function of x satisfying

$y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}}$ where k is a constant and

$y\left( {{1 \over 2}} \right) = - {1 \over 4}$. Then ${{dy} \over {dx}}$ at x = ${1 \over 2}$, is equal to :