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How do you find the derivative of $y = cot x (csc x)$ $y = cotx(cscx)$ ?

1 Answer

Shwetank Mauria

Aug 3, 2016

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

Explanation:

Product rule states if $f (x) = g (x) h (x)$ $f(x)=g(x)h(x)$

then $\frac{d f}{d x} = \frac{d g}{d x} \times h (x) + \frac{d h}{d x} \times g (x)$ $(df)/(dx)=(dg)/(dx)xxh(x)+(dh)/(dx)xxg(x)$

Hence as $y = cot x (csc x)$ $y=cotx(cscx)$

$\frac{d y}{d x} = \frac{d}{d x} (cot x) \times csc x + \frac{d}{d x} (csc x) \times cot x$ $(dy)/(dx)=d/(dx)(cotx)xxcscx+d/(dx)(cscx)xxcotx$

= $(- {csc}^{2} x) \times csc x + (- csc x cot x) \times cot x$ $(-csc^2x)xxcscx+(-cscxcotx)xxcotx$

= $- {csc}^{2} x \times csc x - csc \times x {cot}^{2} x$ $-csc^2x xxcscx-cscxxxcot^2x$

= $- csc x ({csc}^{2} x + {cot}^{2} x)$ $-cscx(csc^2x+cot^2x)$

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