Find f'(x) of this ?

enter image source here

1 Answer
Mar 18, 2017

f'(x)=cosxcotx-cscx-cotxcscx+12/x^5

Explanation:

f(x)=sinxcotx+cscx-3/x^4

Hence f'(x)=d/(dx)(sinxcotx)+d/(dx)cscx-d/(dx)(3/x^4)

= (cosxcotx+sinx xx(-csc^2x))-cotxcscx-3xx(-4)xx1/x^5

= cosxcotx-cscx-cotxcscx+12/x^5