Question

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

Answer 1

`\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}}`

Answer 2

`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))`