Another log derivitive

[sigma]

I have this question:


\(\displaystyle \
\L\
\begin{array}{l}
y = \frac{{2^x }}{{\ln (x^4 + x^2 )}} \\
y' = \frac{{x\ln 2}}{{\frac{1}{{x^4 + x^2 }}(4x^3 + 2x)}} \\
\end{array}
\\)

But I get stuck after that for the numerator. I know the derivitive for\(\displaystyle \
\L\
\ln x
\
\\) is \(\displaystyle \
\L\
\frac{1}{x}
\\)but what is the derivitive for \(\displaystyle \L\ln2\\)? Its not \(\displaystyle \L\frac{1}{2}\\) is it?

 

[royhaas]

The derivative of a constant is zero.

 

[tkhunny]

How did you get that expression for y'? It doesn't look very close.

What rules are you using? The Quotient Rule doesn't seem to have come into play.

 

[soroban]

Hello, sigma!

Ever hear of the Quotient Formula?

\(\displaystyle \L y\:=\:\frac{2^x }{\ln(x^4\,+\,x^2)}\)

\(\displaystyle y' = \frac{x\ln2}{\frac{1}{x^4 + x^2 }(4x^3 + 2x)}\;\;\) . . . awful!

It starts: \(\displaystyle \L\:\frac{\ln(x^4\,+\,x^2)\,\cdot2^x\cdot\,\ln x \;-\;2^x\cdot\,\frac{4x^3\,+\,2x}{x^4\,+\,x^2}}{[\ln(x^4\,+\,x^2)]^2}\)

 

[tkhunny]

* ln(2)?