Question

Find derivative of cot square root x by first principle

Answer 1

- [cosec^2 (√x)] / 2√x

• d (cot√x)/dx  = lim h→0 [cot√(x+h) - cot√x] / h

•                      = lim h→0 [ { 1/tan(√x+h) - 1/tan(√x) } ] / h

which is an indeterminate form of type 0/0

so, we apply L'Hospital's rule,

•    lim h→0 [ { 1/tan(√x+h) - 1/tan(√x) } ] / h
• =  lim h→0 d/dh [ { 1/tan(√x+h) - 1/tan(√x) } ] / (dh/dh)
• =  lim h→0 (d/dh)  { 1/tan(√x+h) - 1/tan(√x) }
• =  -1 / { 2sin√(h+x)^2 * √(x+h) }

thus,

•    lim h→0 [ { 1/tan(√x+h) - 1/tan(√x) } ] / h

• =  lim h→0 [-1 / { 2sin√(h+x)^2 * √(x+h) } ]

• =  -1/2 lim h→0 [ 1 / { sin√(h+x)^2 * √(x+h) } ]

• =  -1 / 2 [ √{lim h→0(h+x) } { lim h→0 sin (h+x)^2 } ]

• =  -1 / 2√x sin [ √{lim h→0(h+x)^2 } ]

• =  - [cosec^2 (√x)] / 2√x