differentiate and do not simplify:

Y=e^ -sq.rt x + e^3 - 2^(3x-1)

finally, a name

http://www.jiskha.com/display.cgi?id=1271798615

Y=e^ -√x + e^3 - 2^(3x-1)
Y=e^ -x^(1/2) + e^3 - 2^(3x-1)
Y' = e^ -x^(1/2)(-1/2)x^(-1/2) + 0 - ln2(2^(3x+1))(3)

y = e^-(x^.5) + e^3 -2^(3x-1)

.5x^(-.5) e^-(x^.5) + 0 - 3[2^(3x-1)]*ln2

To differentiate the function Y = e^(-√x) + e^3 - 2^(3x-1), you will need to apply the derivative rules for each term separately.

Let's differentiate each term step by step:

1. Differentiating the first term, Y = e^(-√x):
To differentiate e^(-√x), we use the chain rule. Let's define u = -√x and apply the chain rule:

dy/dx = d(e^u)/dx = (d(e^u)/du) * (du/dx)

du/dx = d(-√x)/dx = -1/(2√x)

d(e^u)/du = e^u

Now, we can substitute these values back into the equation:

dy/dx = e^u * du/dx = e^(-√x) * (-1/(2√x)) = -e^(-√x) / (2√x)

2. Differentiating the second term, Y = e^3:
The derivative of a constant (e^3 in this case) is zero. So, the differential of this term is 0.

3. Differentiating the third term, Y = 2^(3x-1):
To differentiate 2^(3x-1), we use the chain rule. Let's define u = 3x-1 and apply the chain rule:

dy/dx = d(2^u)/dx = (d(2^u)/du) * (du/dx)

du/dx = d(3x-1)/dx = 3

d(2^u)/du = ln(2) * 2^u

Now, we can substitute these values back into the equation:

dy/dx = ln(2) * 2^u * du/dx = ln(2) * 2^(3x-1) * 3 = 3ln(2) * 2^(3x-1)

Finally, we can put all the differentiated terms together to get the overall derivative of Y:

dy/dx = -e^(-√x) / (2√x) + 3ln(2) * 2^(3x-1)

Thus, the derivative of Y = e^(-√x) + e^3 - 2^(3x-1) is dy/dx = -e^(-√x) / (2√x) + 3ln(2) * 2^(3x-1).