How do i find the derivative for this

square root of x(x+2)e^x

ive tried it and i failed. i tried spliting them up into f(x) and g(x)
im suppoused to use the product rule

best explanation gets best answer thanks

d/dx [x(x+2)] = (x+2) + x = 2 (x+1)
d/dx [e^x] = e^x
d/dx [x(x+2)*e^x] = 2(x+1)*e^x + x(x+2)*e^x = {2x+2+x^2+2x} e^x = {x^2+4x+2}*e^x

d/dx [ sqrt( x(x+2)*e^x ) ] = (1/2) (1/sqrt( x(x+2)*e^x )) . d/dx [x(x+2)*e^x]
.....................................= (1/2) (1/sqrt( x(x+2)*e^x )) . {x^2+4x+2}*e^x
.....................................= (1/2) (1/sqrt( x(x+2)) ) . (x+2)^2 * e^((1/2)x)

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&…
According to the change you made that only the first x is under the square root. The changes are little as follows:
d/dx [√x(x+2)] = (x+2)/(2√x) + √x
d/dx [e^x] = e^x

d/dx [ √x(x+2)*e^x ] = {(x+2)/(2√x) + √x}e^x

Answer is: {(x+2)/(2√x) + √x}e^x

f(x) = √(x(x + 2)e^x)

I'm assuming everything you wrote is under a radical.

f '(x) = 1/2 ((x(x + 2)e^x)^(-1/2)) ∙ [(x + 2)e^x + xe^x + x(x + 2)e^x]
= e^x [x + 2 + x + x(x + 2)] / (2√(x(x + 2)e^x))
= e^x [2x + 2 + x² + 2x] / (2√(x(x + 2)e^x))
= e^x [x² + 4x + 2] / (2√(x(x + 2)e^x))

- - - - - - - - - - - -
Or maybe you meant
f(x) = √x ∙ (x + 2) ∙ e^x

In which case
f '(x) = ((x + 2)e^x)/(2√x) + √x e^x + √x(x + 2)e^x

If g(x) = a(x) * b(x) * c(x)

g'(x)=a'(x) * b(x) * c(x) + a(x) * b'(x) * c(x) + a(x) * b(x) * c'(x)

In your question, f(x) = g(x) ^ (1/2)

So

f'(x) = 1/2 * g(x) ^ (-1/2) * g'(x)

Here g(x) = x (x+2) e^x

So

g'(x) = 1 * (x+2) e^x + x * 1 * e^x + x (x+2) * e^x
= (x+2) e^x + x e^x + x (x+2) e^x

Now you know what's f'(x)

I hope it's helpful :-)

Let,
y = sqrt[x(x+2)e^x] = √x*√(x+2)*√(e^x) = √x*√(x+2)*e^x/2

y' = √(x+2)*e^x/2 / 2√x +√x*e^x/2 / 2√(x+2) +√x*√(x+2)*e^x/2 / 2
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