Differentiate and do not simplify:

Y=x^2 * 3^2x

Y=uv

u=x^2 u'=2x
v=3^2x or v=9^x v'=9^x * ln(9)

y'=uv' + vu'

To differentiate the expression Y = x^2 * 3^(2x) without simplifying, we need to apply the product rule and the chain rule.

The product rule states that if we have two functions u(x) and v(x), then the derivative of their multiplication is given by:
(d/dx)(u(x) * v(x)) = u'(x)v(x) + u(x)v'(x)

In this case, u(x) = x^2 and v(x) = 3^(2x). Let's first find the derivatives of u(x) and v(x):

u'(x) = d/dx (x^2) = 2x

Now, to find the derivative of v(x), we need to use the chain rule because we have a function of the form f(g(x)), with f(x) = 3^x and g(x) = 2x.

Let's define a new function y = 2x. Using the chain rule, we can find the derivative of v(x) as follows:

dv/dx = d/dx (3^(2x)) = d/dy (3^y) * dy/dx (using the chain rule)

To find dy/dx, we differentiate y = 2x:

dy/dx = d/dx (2x) = 2

Next, we differentiate f(x) = 3^x with respect to y:

d/dy (3^y) = ln(3) * 3^y (using the derivative of exponential function rule)

Now we can substitute these values back into our derivative formula:

dv/dx = ln(3) * 3^(2x) * 2

Finally, we can find the derivative of the expression Y = x^2 * 3^(2x) by applying the product rule:

dY/dx = u'(x)v(x) + u(x)v'(x)
= (2x)(3^(2x)) + (x^2)(ln(3) * 3^(2x) * 2)
= 2x * 3^(2x) + 2ln(3) * x^2 * 3^(2x)

Therefore, the differentiated expression is dY/dx = 2x * 3^(2x) + 2ln(3) * x^2 * 3^(2x).