Question

Find \(y'\)if \(y = \ln \left( {{x^2} + {y^2}} \right)\).

Short answer

The answer is \(y' = \frac{{2x}}{{{x^2} + {y^2} - 2y}}\).

Step-by-step solution

Logarithmic Differentiation

For some complicated functions of the form \({a^{f\left( x \right)}}\), \(f{\left( x \right)^{g\left( x \right)}}\) , the usual differentiation will become more complicated. To differentiate these types of function we first takelogarithm and then differentiate them.

For example:

\(y = {a^{f\left( x \right)}}\). Now taking natural log both side we get,

\(\ln y = f\left( x \right)\ln a\).

Differentiating we get,

\(\begin{aligned}\frac{1}{y}\frac{{dy}}{{dx}} &= f'\left( x \right)\ln a\\\frac{{dy}}{{dx}} &= yf'\left( x \right)\ln a\end{aligned}\)

Again, for the function \(y = f{\left( x \right)^{g\left( x \right)}}\), taking logarithm we get,

\(\ln y = g\left( x \right)\ln \left( {f\left( x \right)} \right)\)

Now differentiating,

\(\begin{aligned}\frac{1}{y}\frac{{dy}}{{dx}} &= g'\left( x \right)\ln \left( {f\left( x \right)} \right) + \frac{{g\left( x \right)}}{{f\left( x \right)}}f'\left( x \right)\\\frac{{dy}}{{dx}} &= y\left( {g'\left( x \right)\ln \left( {f\left( x \right)} \right) + \frac{{g\left( x \right)}}{{f\left( x \right)}}f'\left( x \right)} \right)\end{aligned}\)

Finding the derivative of the function

Given that \(y = \ln \left( {{x^2} + {y^2}} \right)\).

Now differentiating with respect to \(x\) we get,

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

Hence finally we get \(y' = \frac{{2x}}{{{x^2} + {y^2} - 2y}}\).