How do you find the derivative of #sqrt(x+1)#?

2 Answers

Answer:

#\frac{d}{dx}\sqrt{x+1}=1/{2\sqrt{x+1}}#

Explanation:

Differentiating w.r.t. #x# by using chain rule as follows

#\frac{d}{dx}\sqrt{x+1}#

#=\frac{d}{dx}(x+1)^{1/2}#

#=1/2(x+1)^{1/2-1}\frac{d}{dx}(x+1)#

#=1/2(x+1)^{-1/2}(1)#

#=1/2(x+1)^{-1/2}#

#=1/{2\sqrt{x+1}}#

Jul 12, 2018

Answer:

#1/(2sqrt(x+1))#

Explanation:

#"differentiate using the "color(blue)"chain rule"#

#"given "y=f(g(x))" then"#

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#"here "y=sqrt(x+1)=(x+1)^(1/2)#

#dy/dx=1/2(x+1)^(-1/2)xxd/dx(x+1)#

#color(white)(dy/dx)=1/(2sqrt(x+1))#