  ## Logarithmic Differentiation

Suppose we wish to find $\displaystyle{\frac{dy}{dx}}$, where $\displaystyle{y = \frac{\sqrt{x+1}}{(x+2)^6\sqrt{x+3}}}$.

At first blush, we might think we need to employ the quotient rule, the product rule, and a couple of chain rule applications involving derivatives of powers -- a task that is certianly doable, but likely to be algebraically tedious!

However, there is a simpler way!

For some complicated expressions involving product, quotients, and powers, we can use the properties of logarithms to make the expression more "differentiation-friendly".

Recall that $e^{\ln x} = x$

As such,

$${\large y} = {\large e}^{\left(\displaystyle{\ln \frac{\sqrt{x+1}}{(x+2)^6\sqrt{x+3}}}\right)}$$

However, recalling $\ln AB = \ln A + \ln B$ and $\ln A/B = \ln A - \ln B$ under the appropriate conditions, we can break the exponent up into the sum/difference of some smaller pieces:

$${\large y} = {\large e}^{\left(\displaystyle{\ln \sqrt{x+1} - \ln (x+2)^6 - \ln \sqrt{x+3}}\right)}$$

Then, recalling that $\ln A^p = p \ln A$ we can do even more to simplify these pieces.

$${\large y} = {\large e}^{\left(\displaystyle{\frac{1}{4}\ln (x+1) - 6\ln (x+2) - \frac{1}{2}\ln (x+3)}\right)}$$

Now, we can use the chain rule to find $dy/dx$:

$$\frac{dy}{dx} = {\large e}^{\left(\displaystyle{\frac{1}{4}\ln (x+1) - 6\ln (x+2) - \frac{1}{2}\ln (x+3)}\right)} \cdot \left(\frac{1}{4(x+1)} - \frac{6}{x+2} - \frac{1}{2(x+3)} \right)$$

As one final simplification, note that the power of $e$ above was just another way of writing the original $y$, and substitute this out for the formula given to us for $y$:

$$\frac{dy}{dx} = \frac{\sqrt{x+1}}{(x+2)^6\sqrt{x+3}} \cdot \left(\frac{1}{4(x+1)} - \frac{6}{x+2} - \frac{1}{2(x+3)} \right)$$

Ta-Da! (That's a whole lot easier than using the quotient rule, product rule, and chain rule all together, don't you think?)

This process of rewriting $y$ as $e$ to a logarithm, and then using the properties of logarithms to simplify the exponent before differentiating (and finally making one last substitution) is called logarithmic differentiation and can be a real time saver under the right circumstances!