Implicit Differentiation: Additional Notes

Implicit differentiation: additional notes

For a constant base \(a>0\), \(\frac{\mathrm d}{\mathrm dx}(a^x)=a^x\ln a\); for a constant exponent, \(\frac{\mathrm d}{\mathrm dx}(x^n)=nx^{n-1}\). When both the base and exponent vary, logarithmic differentiation is useful.

For \(x>0\), with \(a>0\), \(a\ne1\), and \(b>0\) where used, the related logarithmic rules are \(\frac{\mathrm d}{\mathrm dx}(\ln x)=\frac1x\), \(\frac{\mathrm d}{\mathrm dx}(\log_a x)=\frac1{x\ln a}\), and \(\log_a b=\frac{\ln b}{\ln a}\).

For the example \(y=x^x\) on \(x>0\), take natural logarithms to obtain \(\ln y=x\ln x\).

Differentiating implicitly gives \(\frac1y\frac{\mathrm dy}{\mathrm dx}=x\left(\frac1x\right)+\ln x=1+\ln x\).

Thus \(\frac{\mathrm dy}{\mathrm dx}=y(1+\ln x)=x^x(1+\ln x)\).