Differentiating this function could be done with a product rule and a quotient rule. However, that would be a fairly messy process. We can simplify things somewhat by taking logarithms of both sides.

\[\ln y = \ln \left( {\frac{{{x^5}}}{{\left( {1 - 10x} \right)\sqrt {{x^2} + 2} }}} \right)\]

Of course, this isn’t really simpler. What we need to do is use the properties of logarithms to expand the right side as follows.

\[\begin{align*}\ln y & = \ln \left( {{x^5}} \right) - \ln \left( {\left( {1 - 10x} \right)\sqrt {{x^2} + 2} } \right)\\ \ln y & = \ln \left( {{x^5}} \right) - \ln \left( {1 - 10x} \right) - \ln \left( {\sqrt {{x^2} + 2} } \right)\end{align*}\]

This doesn’t look all that simple. However, the differentiation process will be simpler. What we need to do at this point is differentiate both sides with respect to \(x\). Note that this is really implicit differentiation.

\[\begin{align*}\frac{{y'}}{y} & = \frac{{5{x^4}}}{{{x^5}}} - \frac{{ - 10}}{{1 - 10x}} - \frac{{\frac{1}{2}{{\left( {{x^2} + 2} \right)}^{ - \frac{1}{2}}}\left( {2x} \right)}}{{{{\left( {{x^2} + 2} \right)}^{\frac{1}{2}}}}}\\ \frac{{y'}}{y} & = \frac{5}{x} + \frac{{10}}{{1 - 10x}} - \frac{x}{{{x^2} + 2}}\end{align*}\]

To finish the problem all that we need to do is multiply both sides by \(y\) and the plug in for \(y\) since we do know what that is.

\[\begin{align*}y' & = y\left( {\frac{5}{x} + \frac{{10}}{{1 - 10x}} - \frac{x}{{{x^2} + 2}}} \right)\\ & = \frac{{{x^5}}}{{\left( {1 - 10x} \right)\sqrt {{x^2} + 2} }}\left( {\frac{5}{x} + \frac{{10}}{{1 - 10x}} - \frac{x}{{{x^2} + 2}}} \right)\end{align*}\]

Depending upon the person, doing this would probably be slightly easier than doing both the product and quotient rule. The answer is almost definitely simpler than what we would have gotten using the product and quotient rule.

We’ve seen two functions similar to this at this point.

\[\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\hspace{0.5in}\frac{d}{{dx}}\left( {{a^x}} \right) = {a^x}\ln a\]

Neither of these two will work here since both require either the base or the exponent to be a constant. In this case both the base and the exponent are variables and so we have no way to differentiate this function using only known rules from previous sections.

With logarithmic differentiation we can do this however. First take the logarithm of both sides as we did in the first example and use the logarithm properties to simplify things a little.

\[\begin{align*}\ln y & = \ln {x^x}\\ \ln y & = x\ln x\end{align*}\]

Differentiate both sides using implicit differentiation.

\[\frac{{y'}}{y} = \ln x + x\left( {\frac{1}{x}} \right) = \ln x + 1\]

As with the first example multiply by \(y\) and substitute back in for \(y\).

\[\begin{align*}y' & = y\left( {1 + \ln x} \right)\\ & = {x^x}\left( {1 + \ln x} \right)\end{align*}\]

Now, this looks much more complicated than the previous example, but is in fact only slightly more complicated. The process is pretty much identical, so we first take the log of both sides and then simplify the right side.

\[\ln y = \ln \left[ {{{\left( {1 - 3x} \right)}^{\cos \left( x \right)}}} \right] = \cos \left( x \right)\ln \left( {1 - 3x} \right)\]

Next, do some implicit differentiation.

\[\frac{{y'}}{y} = - \sin \left( x \right)\ln \left( {1 - 3x} \right) + \cos \left( x \right)\frac{{ - 3}}{{1 - 3x}} = - \sin \left( x \right)\ln \left( {1 - 3x} \right) - \cos \left( x \right)\frac{3}{{1 - 3x}}\]

Finally, solve for \(y'\) and substitute back in for \(y\).

\[\begin{align*}y' & = - y\left( {\sin \left( x \right)\ln \left( {1 - 3x} \right) + \cos \left( x \right)\frac{3}{{1 - 3x}}} \right)\\ & = - {\left( {1 - 3x} \right)^{\cos \left( x \right)}}\left( {\sin \left( x \right)\ln \left( {1 - 3x} \right) + \cos \left( x \right)\frac{3}{{1 - 3x}}} \right)\end{align*}\]

A messy answer but there it is.