Question #548e9

1 Answer
Mar 19, 2017

d/dxcotx = -1/sin^2x

Explanation:

By definition the derivative of f(x) is:

lim_(h->0) (f(x+h)-f(x))/h

For f(x) = cotx:

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

Now use the trigonometric identity:

cot(x+h) = cos(x+h)/sin(x+h) = ( cosxcosh-sinxsinh)/(cosxsinh+sinxcosh) = ( (cosxcosh)/(sinxsinh)-1)/((cosxsinh+sinxcosh)/(sinxsinh)) = (cotxcoth -1) / (cotx+cot h)

and we have:

d/dxcotx = lim_(h->0) ((cotxcoth -1) / (cotx+cot h)-cot(x))/h

d/dxcotx = lim_(h->0) (cotxcoth -1 -cot(x)(cotx+cot h))/(h(cotx+cot h))

d/dxcotx = lim_(h->0) (cancel(cotxcoth) -1 -cot^2x-cancel(cotxcot h))/(h(cotx+cot h))

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

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

evaluate the limit by expading the denominator:

lim_(h->0) 1/(h(cotx+cot h)) = lim_(h->0) 1/(h(cosx/sinx+cos h/sinh))

lim_(h->0) 1/(h(cotx+cot h)) = lim_(h->0) 1/(hcosx/sinx+(cos h*(h/sinh))

And as lim_(h->0) sinh/h = 1:

lim_(h->0) 1/(h(cotx+cot h)) = lim_(h->0) 1/(0+(1*1)) = 1

so:

d/dxcotx = - ( 1 +cot^2x)

and as:

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

we can conclude:

d/dxcotx = -1/sin^2x