How do you find the derivative of $y = cot5x + csc5x$ ?

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Answer 1 · 2018-05-13T00:23:46

$-5csc 5x (cosec 5x-cot 5x)or -5 csc 5x × tan 5x/2$

Given $y=cot 5x+csc 5x$
$d/dx(cot x)=-csc^2 x$
$d/dx(csc x)=-csc x cot x$

Now, By chain rule,
$d/dx(cot 5x)=-5csc^2 5x$
$d/dx(csc 5x)=-5csc 5x cot 5x$

Adding, we get,
$-5csc 5x (cosec 5x-cot 5x)$ which is the answer.

Further simplifying,
$csc 5x-cot 5x " as " tan (5x/2)$ by the identity $csc x-cot x =tan (x/2)$
We get the answer as $-5 csc 5x × tan 5x/2$

How do you find the derivative of y = cot5x + csc5x y = cot5x + csc5x ?