Find derivative of cscx using first principal or definition of derivative?

1 Answer
Feb 11, 2018

d/dx(csc x)=-cotxcscx

Explanation:

d/dx (csc x)

= d/dx (1/(sinx))

Using the limit definition of the derivative:
Let f(x) = cscx = 1/sinx

= lim_(h->0) frac{f(x+h)-h(x)}{h}

= lim_(h->0) frac{csc(x+h) - cscx}{h}

= lim_(h->0) frac{(1/sin(x+h) - 1/sinx)}{h}

= lim_(h->0) (sinx-sin(x+h))/(hsin(x+h)sinx)

= lim_(h->0) (2cos((x+x+h)/2)sin((x-x-h)/2))/(hsin(x+h)sinx)

= lim_(h->0) (-2cos(x+h/2)sin(h/2))/(hsin(x+h)sinx)

= lim_(h->0) (-2cos(x+h/2))/(hsin(x+h)sinx)xxsin(h/2)/(h/2)xxh/2

= lim_(h->0) (-cos(x+h/2))/(sin(x+h)sinx)xxsin(h/2)/(h/2)

= lim_(h->0) (-cos(x+h/2))/(sin(x+h)sinx)xxlim_(h->0)sin(h/2)/(h/2)

= -cosx/(sinxsinx)xx1

= -cotxcscx