Question 13 Marks
Find the derivative of function cosec x cot x.
Answer
View full question & answer→Here f (x) = cosec x cot x
$\therefore $f'(x) = $\frac{d}{{dx}}$ [cosec x cot x]
= cosec x $\frac{d}{{dx}}$ (cot x) + cot x $\frac{d}{{dx}}$ (cosec x)
$=\operatorname{cosec} x \cdot-\operatorname{cosec}^2 x+\cot x \cdot-\operatorname{cosec} x \cot x$
$=-\operatorname{cosec}^3 x-\operatorname{cosec} x \cot ^2 x$
$\therefore $f'(x) = $\frac{d}{{dx}}$ [cosec x cot x]
= cosec x $\frac{d}{{dx}}$ (cot x) + cot x $\frac{d}{{dx}}$ (cosec x)
$=\operatorname{cosec} x \cdot-\operatorname{cosec}^2 x+\cot x \cdot-\operatorname{cosec} x \cot x$
$=-\operatorname{cosec}^3 x-\operatorname{cosec} x \cot ^2 x$