How do you differentiate #sqrt(16-x^2)#?

1 Answer
seol
Nov 12, 2016

#d/dxsqrt(16-x^2)=-1/sqrt(16-x^2)#

Explanation:

diff'n of #sqrt(x)# is #d/dxx^(1/2)#
apply #n-1# rule: #1/2x^(1/2-1)\rightarrow1/2x^(-1/2)#

therefore...
#d/dxsqrt(16-x^2)#

apply chain rule
#1/2(16-x^2)^(-1/2)\timesd/dx(16-x^2)#

differentiate
#(1/(2\times(16-x^2)^(1/2)))\times-2x#

simplify:
#-1/sqrt(16-x^2)#

Related questions
See all questions in Chain Rule
Impact of this question
5817 views around the world
You can reuse this answer
Creative Commons License