Question #c60ff

1 Answer
Feb 14, 2017

d/dx (sqrt cot x ) = -csc^2x/(2sqrt(cotx))

Explanation:

d/dx (sqrt cot x ) = lim_(h->0) (sqrt cot(x+h) - sqrt(cot x))/h

Rationalize the numerator using the identity (a^2-b^2) = (a+b)(a-b)

d/dx (sqrt cot x ) = lim_(h->0) ((sqrt cot(x+h) - sqrt(cot x))/h)( (sqrt cot(x+h) + sqrt(cot x))/(sqrt cot(x+h) + sqrt(cot x)))

d/dx (sqrt cot x ) = lim_(h->0) ( cot(x+h) - (cot x))/(h (sqrt cot(x+h) + sqrt(cot x))

Now use the trigonometric formulas for the sum of angles:

cot(x+h) = cos(x+h)/sin(x+h) = (cosxcos h-sinxsin h )/ ( cosx sin h + sinx cos h )

divide numerator and denominator by sinx sin h

cot(x+h) = (cotxcot h-1 )/ ( cotx +cot h )

end evaluate the difference:

cot(x+h) - cotx = (cotxcot h-1 )/ ( cotx +cot h ) - cotx

cot(x+h) - cotx = ((cotxcot h-1 ) -cotx( cotx +cot h ))/ ( cotx +cot h )

cot(x+h) - cotx = ((cancel(cotxcot h) -1 -cot^2x- cancel(cotx cot h )))/ ( cotx +cot h )

cot(x+h) - cotx =- ( 1 +cot^2x)/ ( cotx +cot h )

and as:

1+cot^2x = 1+ cos^2x/sin^2x = (sin^2x+cos^2x)/sin^2x = 1/sin^2x = csc^2x

we have finally:

cot(x+h) - cotx = - csc^2x/(cotx +cot h )

So:

d/dx (sqrt cot x ) = lim_(h->0) ( - csc^2x/(cotx +cot h ))1/(h (sqrt cot(x+h) + sqrt(cot x))

d/dx (sqrt cot x ) = lim_(h->0) (- csc^2x/ (sqrt cot(x+h) + sqrt(cot x)))(1/(h(cotx +coth)))

Now note that:

lim_(h->0) 1/(h(cotx +coth)) = lim_(h->0) 1/(hcotx + cosh/(sinh/h)) =1

and we can conclude:

d/dx (sqrt cot x ) = -csc^2x/(2sqrt(cotx))