Differentiation, derivatives and Taylor approximations: Differentiating exponential and logarithmic functions
Logarithmic differentiation
Suppose that you want to differentiate a function of the form \[y(x)=\bigl(f(x)\bigr)^{g(x)}\] with \(f(x)>0\). This can be accomplished by rewriting the function as \[y(x)=e^{g(x)\cdot \ln\bigl(f(x)\bigr)}\] and applying the chain rule and product rule for differentiation.
Note that you get the result of rewriting of the function by applying the natural logarithm. This gives \(\ln\bigl(y(x)\bigr)=\ln\left( \bigl(f(x)\bigr)^{g(x)}\right)=g(x)\cdot \ln\bigl(f(x)\bigr)\). Apply the exponential function on both sides of the equation and simplify the left-hand side.
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But finding the derivative \(y'(x)\) can also be done via the method known under the name logarithmic differentiation. Here you first apply the natural logarithm on either side of the expression; you will then \[\ln\bigl(y(x)\bigr)=g(x)\cdot \ln\bigl(f(x)\bigr)\tiny.\] Hereafter you compute the derivatives of the expressions on the left- and right-hand side of the equal sign; this leads via the rules for differentiation to\[\frac{y'(x)}{y(x)}=g'(x)\cdot \ln\bigl(f(x)\bigr)+ g'(x)\cdot \frac{f'(x)}{f(x)}\tiny.\] So \[\begin{aligned}y'(x)&=y(x)\cdot\left(g'(x)\cdot \ln\bigl(f(x)\bigr)+ g'(x)\cdot \frac{f'(x)}{f(x)}\right)\\ \\ &= \bigl(f(x)\bigr)^{g(x)}\cdot \left(g'(x)\cdot \ln\bigl(f(x)\bigr)+ g'(x)\cdot \frac{f'(x)}{f(x)}\right)\tiny.\end{aligned}\]
More instructive is the following example of logarithmic differentiation.
Example of logarithmic differentiate
Compute \(y'(1)\) for \(y(x)=\sqrt{x(x+1)(x+2)}\).
Solution
Apply the natural logarithm on both sides of \(y(x)=\sqrt{x(x+1)(x+2)}\) and simplificy. This gives\[\begin{aligned}\ln\bigl(y(x)\bigr)&= \ln\left(\sqrt{x(x+1)(x+2)}\right) \\ &= \ln\Bigl(\bigl(x(x+1)(x+2)\bigr)^{\frac{1}{2}}\Bigr)\\ &=\frac{1}{2}\cdot \ln\bigl(x(x+1)(x+2)\bigr)\\ &= \frac{1}{2}\cdot \bigl(\ln(x)+\ln(x+1)+\ln(x+2)\bigr)\tiny.\end{aligned}\] Differentiate the left- and right-hand side: \[\frac{y'(x)}{y(x)} = \frac{1}{2}\cdot \left(\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x+2}\right)\tiny.\] For \(x=1\) You get: \[y(1)=\sqrt{6},\quad\text{and}\quad y'(1) = \tfrac{1}{2}\sqrt{6}\cdot \left(1+\frac{1}{2}+\frac{1}{3}\right)=\frac{11}{12}\sqrt{6}\tiny.\]
Mathcentre video
Differentiation by Taking Logarithms (19:17)
