Derivade: Y=(4e^x)(x^3x)If you can please help me with: Y=(e^2)+(x^e)+(e^square root of x)

We have to find the derivative of Y=
(4e^x)(x^3x)

Now using the product rule we
get

Y' = (4e^x)'*(x^3x) + (4e^x)*(x^3x)'
...(1)

Now (4e^x)' = 4e^x

To
find the derivative of x^3x , let z = x^3x

Take the
logarithm on both the sides

=> ln z = 3x*ln
x

Take the derivative of both the
sides

=> z'*(1/z) = 3x*(1/x)+3*ln
x

Substituting z=
x^3x

=> z'(1/ x^3x) = 3 + 3 ln
x

=> z' = 3x^3x + 3x^3x * ln
x

=> (x^3x)' = 3x^3x + 3x^3x * ln
x

Now substitute these values in
(1)

Y' = (4e^x)*(x^3x) + (4e^x)*(3x^3x + 3x^3x * ln
x)

=> Y' = (4e^x)*(x^3x)*[1 + (3 + 3 * ln
x)]

=> Y' = (4e^x)*(x^3x)*(4 + 3 * ln
x)

Therefore the derivative of (4e^x)(x^3x)
is (4e^x)*(x^3x)*(4 + 3 * ln x)