Question

Find the derivative using first principles? : `cscx`

Answer 1

`d/dx cscx = lim_(h rarr 0) ( 1/sin(x+h) - 1/sin(x) ) / h = -cscx \ cotx`

Explanation:

We seek: `d/dx cscx` from first principles:

We have, using the definition of the derivative that:

`d/dx cscx = lim_(h rarr 0) ( csc(x+h)-csc(x) ) / h`

`\ \ \ \ \ \ \ \ \ \ \ \ \= lim_(h rarr 0) ( 1/sin(x+h) - 1/sin(x) ) / h`

`\ \ \ \ \ \ \ \ \ \ \ \ \= lim_(h rarr 0) ( (sinx - sin(x+h) ) / (sin(x)sin(x+h) ) ) / h`

`\ \ \ \ \ \ \ \ \ \ \ \ \= lim_(h rarr 0) (sinx - sin(x+h) ) / (hsin(x)sin(x+h) )`

Using `sinA-sinB -= 2cos((A+B)/2)cos((A-B)/2)` we have:

`d/dx cscx = lim_(h rarr 0) (2cos((x+x+h)/2)cos((x-x-h)/2) ) / (hsin(x)sin(x+h) )`

`\ \ \ \ \ \ \ \ \ \ \ \ \= lim_(h rarr 0) (2cos((2x+h)/2)cos((-h)/2) ) / (hsin(x)sin(x+h) )`

`\ \ \ \ \ \ \ \ \ \ \ \ \= lim_(h rarr 0) (cos(x+h/2))/(sin(x)sin(x+h)) * (2cos(-h/2) ) / (h)`

`\ \ \ \ \ \ \ \ \ \ \ \ \= lim_(h rarr 0) (cos(x+h/2))/(sin(x)sin(x+h)) * lim_(h rarr 0) (-cos(h/2) ) / (h/2)`

`\ \ \ \ \ \ \ \ \ \ \ \ \= lim_(h rarr 0) (cos(x+h/2))/(sin(x)sin(x+h)) * lim_(theta rarr 0) - (cos(theta) ) / (theta)`

The second limit is a standard calculus limit and its value is unity, so we can now evaluate the limit :

`d/dx cscx = (cos(x+0))/(sin(x)sin(x+0)) * (-1)`

`\ \ \ \ \ \ \ \ \ \ \ \ \= -(cosx)/(sinxsinx)`

`\ \ \ \ \ \ \ \ \ \ \ \ \= -1/sinx * cosx/sinx`

`\ \ \ \ \ \ \ \ \ \ \ \ \= -cscx \ cotx`