Derivade:

Y=(4e^x)(x^3x)

If you can please help me with:

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

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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)**

y = (4e^x)(x^(3x)).

We take the logarithm of both sides:

logy = ln(4e^x) +ln(x^(3x))

logy = ln4 +x +3x* lnx.

Now we differentiate both sides with respect to x:

(1/y) (dy/dx) = {ln4+x+3x*lnx)'.

(1/y)(dy/dx) = (ln4)' +(x)' + (3x*lnx)'.

(1/y)(dy/dx) = 0+1 +(3x)'lnx +3x(lnx)'.

(1/y )(dy/dx) = 1+3lnx +3x/x .

(1/y)(dy/dx) = 4+3lnx.

dy/dx = (4+3lnx)y.

dy/dx = (4+3lnx)(4e^x)(x^3x).

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