Question

20.Find \(\frac{{dy}}{{dx}}\) by implicit differentiation.

20. \(xy = \sqrt {{x^2} + {y^2}} \)

Short answer

The value is \(\frac{{dy}}{{dx}} = \frac{{x - y\sqrt {{x^2} + {y^2}} }}{{x\sqrt {{x^2} + {y^2}} - y}}\).

Step-by-step solution

Differentiate the given equation

The given equation is\(xy = \sqrt {{x^2} + {y^2}} \). Both sides of this equation are differentiated with respect to\(x\)as follows:

\(\begin{aligned}\frac{d}{{dx}}\left( {xy} \right) &= \frac{d}{{dx}}\left( {\sqrt {{x^2} + {y^2}} } \right)\\xy' + y\left( 1 \right) &= \frac{1}{2}{\left( {{x^2} + {y^2}} \right)^{ - \frac{1}{2}}}\left( {2x + 2yy'} \right)\end{aligned}\)

Simplify the resulting equation

Simplify the resulting equation to obtain the expression for\(\frac{{dy}}{{dx}}\), as shown below:

\(\begin{aligned}xy' + y &= \frac{x}{{\sqrt {{x^2} + {y^2}} }} + \frac{y}{{\sqrt {{x^2} + {y^2}} }}y'\\xy' - \frac{y}{{\sqrt {{x^2} + {y^2}} }}y' &= \frac{x}{{\sqrt {{x^2} + {y^2}} }} - y\\\frac{{x\sqrt {{x^2} + {y^2}} - y}}{{\sqrt {{x^2} + {y^2}} }}y' &= \frac{{x - y\sqrt {{x^2} + {y^2}} }}{{\sqrt {{x^2} + {y^2}} }}\\y' &= \frac{{x - y\sqrt {{x^2} + {y^2}} }}{{x\sqrt {{x^2} + {y^2}} - y}}\end{aligned}\)

Thus, the value is \(\frac{{dy}}{{dx}} = \frac{{x - y\sqrt {{x^2} + {y^2}} }}{{x\sqrt {{x^2} + {y^2}} - y}}\).