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

Question

1013 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-11T06:01:00

#d/dx(csc x)=-cotxcscx#

#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#