I am having problems calculating derivatives of squared trigonometric functions.
$$f(x) = \csc^2(4x)$$
Let's see... This is equivalent to
$$f(x) = \csc(4x)\cdot \csc(4x)$$
Now, to get the derivative, we use the product rule:
$$f'(x) = 2 \cdot (-\csc(4x)\cdot\cot(4x)\cdot\csc(4x))$$
But this appears to be wrong when I evaluate this in a calculator. What is the problem with this procedure?
$\endgroup$ 13 Answers
$\begingroup$Hint: $$y=(f(ax))^2$$ $$y'=2f(ax)f'(ax)(a)$$
$\endgroup$ $\begingroup$It is probably easier to write $$f(x)=\frac{1}{\sin^{2}(4x)}$$ so that $$\begin{align} f'(x)&=\frac{-1}{\sin^{4}(4x)}\cdot(\sin^{2}(4x))'\\ &=\frac{-2\sin(4x)\cos(4x)\cdot4}{\sin^{4}(4x)}\\ &=\frac{-8\cos(4x)}{\sin^{3}(4x)}\\ &=-8\cot(4x)\csc^{2}(4x) \end{align}$$
$\endgroup$ $\begingroup$Notice, Applying chain rule as follows $$\frac{d}{dx}(\csc^2 (4x) )=\frac{d}{dx}(\csc (4x) )^2$$$$=2\csc (4x)\frac{d}{dx}(\csc (4x))$$ $$=2\csc(4x)(-\csc(4x)\cot(4x))\frac{d}{dx}(4x)$$$$=2\csc(4x)(-\csc(4x)\cot(4x))(4)$$ $$=\color{red}{-8\csc^2(4x)\cot(4x)}$$
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