Find trig derivative of $y=4x(7x+\cot{7x})^6$ Find trig derivative of $y=4x(7x+\cot{7x})^6$.
I got $y'= 4(7x+\cot{7x})^6 + 168x(7x+\cot{7x})^5 (\cot^2 {7x})$ but I'm not sure I did it right.  
Your help is appreciated (:
 A: Let's just go through it step by step:
$$y=4x(7x+\cot(7x))^6$$
The first step in taking the derivative is using the product rule:
$$y' = (4x)'(7x+\cot(7x))^6+(4x)[(7x+\cot(7x))^6]' \\ = 4(7x+\cot(7x))^6+ 4x[(7x+\cot(7x))^6]'$$
Now we need the derivative of $(7x+\cot(7x))^6$.  Here our next step is the chain rule.  Our outside function is $f(y)=y^6$ and our inside function is $g(x)=7x+\cot(7x)$, so the derivative is just $$[(7x+\cot(7x))^6]' = 6[(7x+\cot(7x))^5]\cdot[7+(\cot(7x))']$$
And finally, we have to do the chain rule one more time to see that $(\cot(7x))'=-7\csc^2(7x)$.
Putting this all together, we've got
$$y' = 4(7x+\cot(7x))^6+ 4x\cdot 6[(7x+\cot(7x))^5]\cdot[7-7\csc^2(7x)]$$
Then you're done (unless you feel like simplifying this, but there's no more calculus to do).
A: use the logarithmic derivative. here is how it works. 
let $$y = 4x(7x + \cot 7x)^6, \ln y = \ln 4 + \ln x + 6 \ln(7x \sin 7x + \cos 7x) - 6 \ln \sin 7x $$ taking the derivative, we get 
$$\begin{align}\frac 1y\frac{dy}{dx} &=\frac 1x+6\left(\frac{7\sin 7x+49x\cos7 x-7\sin7 x}{7x\sin 7x + \cos 7x}- \frac{7\cos 7x}{\sin 7x}\right) \\
&=\frac 1x+42\left(\frac{(\sin 7x+7x\cos 7x-\sin 7x)\sin 7x - \cos 7x(7x \sin 7x+\cos 7x)}{7x\sin 7x + \cos 7x}\right)\\
&=\frac 1x-\frac{ 42\cos^2 7x}{7x\sin 7x + \cos 7x}
\end{align}$$
