How do you find the derivative of $y = cot 5 x + csc 5 x$ $y = cot5x + csc5x$ ?

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How do you find the derivative of $y = cot 5 x + csc 5 x$ $y = cot5x + csc5x$ ?

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Answer 1 · 2018-05-13T00:23:46

$- 5 csc 5 x (cos e c 5 x - cot 5 x) or - 5 csc 5 x \times tan 5 \frac{x}{2}$ $-5csc 5x (cosec 5x-cot 5x)or -5 csc 5x × tan 5x/2$

Explanation:

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

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

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

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

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How do you find the derivative of $y = cot 5 x + csc 5 x$ $y = cot5x + csc5x$ ?

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Explanation:

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

1 Answer

Explanation:

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How do you find the derivative of $y = cot 5 x + csc 5 x$ $y = cot5x + csc5x$ ?