# Find the derivate using differentiation rules. `f(x)= x sqrt(x+1)`I broke it down like this: (just not sure if that was right) x(x+1)^1/2 x*1/2(x+1)^-1/2 + (x+1)^1/2 *x x/[2(x+1)^1/2] + (x+1)^1/2...

Find the derivate using differentiation rules. `f(x)= x sqrt(x+1)`

I broke it down like this: (just not sure if that was right)

x(x+1)^1/2

x*1/2(x+1)^-1/2 + (x+1)^1/2 *x

x/[2(x+1)^1/2] + (x+1)^1/2 *x

x/[2(x+1)^1/2] + [(x+1)^1/2 *x]/1 * [2(x+1)^1/2]/[2(x+1)^1/2]

[2x(x+1)]/[2(x+1)^1/2]

(3x+2)/[2(x+1)^1/2]

*print*Print*list*Cite

`f(x) = xsqrt(x+1)`

To find f '(x), apply the product rule of derivatives which is `(u*v)' = uv' + vu'` .

So, let:

`u=x` and `v=sqrt(x+1)=(x+1)^(1/2)`

Then, take the derivative of u and v. Use the power formula of derivative which is `(u^n)' = n u^(n-1) u'` .

`u'=1` and `v'=1/2 *(x+1)^(-1/2)*1 = 1/(2(x+1)^(1/2))`

Substitute u,v, u' and v' to the formula of product rule of derivatives.

`f '(x) = x*1/(2(x+1)^(1/2)) + (x+1)^(1/2)*1`

`f'(x) = x/(2(x+1)^(1/2)) + (x+1)^(1/2)`

Then, express left side as fractions with same denominator.

`f'(x) = x/(2(x+1)^(1/2)) + ((x+1)^(1/2)*2(x+1)^(1/2))/(2(x+1)^(1/2))`

`f'(x)= x/(2(x+1)^(1/2)) + (2(x+1))/(2(x+1)^(1/2))`

`f'(x)= x/(2(x+1)^(1/2)) + (2x+2)/(2(x+1)^(1/2))`

`f'(x) = (3x+2)/(2(x+1)^(1/2))`

`f'(x)= (3x+2)/(2sqrt(x+1))`

**Hence the derivative of `f(x) = xsqrt(x+1)` is `f'(x)=(3x+2)/(2sqrt(x+1))` .**