differentiate and do not simplify:
Y=e^ -sq.rt x + e^3 - 2^(3x-1)
finally, a name
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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).