Differentiate and do not simplify:
Y=e^-sq.rt.x + e^3 - 2^(3x-1)
To differentiate the expression Y = e^(-√x) + e^3 - 2^(3x-1), you can use the rules of differentiation for each term separately. Let's break it down step by step:
1. Differentiating e^(-√x):
The derivative of e^(-√x) with respect to x can be found using the chain rule. The chain rule states that if you have a function g(f(x)), then its derivative is g'(f(x)) * f'(x). In this case, g(x) = e^x and f(x) = -√x. Therefore, the derivative of e^(-√x) with respect to x is:
(d/dx) [e^(-√x)] = e^(-√x) * (-1/2√x) * (d/dx)(-√x) = e^(-√x) * (-1/2√x) * (-1/2√x)
Simplifying the above expression gives:
(d/dx) [e^(-√x)] = e^(-√x) / (2x)
2. Differentiating e^3:
The derivative of a constant is always zero. In this case, the constant is e^3. Therefore,
(d/dx) [e^3] = 0
3. Differentiating 2^(3x-1):
To differentiate 2^(3x-1), we can use the chain rule again. Let u = 3x - 1 and y = 2^u. The derivative of y with respect to x is:
(d/dx) [2^(3x-1)] = (d/dx) [2^u] = (d/dx) [2^(3x-1)] = (d/dx) [2^u] = (d/dx) [e^(u ln 2)] = (d/dx) [e^(u ln 2)] = (d/du) [e^(u ln 2)] * (du/dx) = e^(u ln 2) * (d/du)[u ln 2] * (du/dx)
Simplifying and plugging in the value of u = 3x - 1:
(d/dx) [2^(3x-1)] = 2^(3x-1) * ln 2 * (3) = 3(2^(3x-1))ln2
Now, combining all the derivatives:
(d/dx) [Y] = (d/dx) [e^(-√x)] + (d/dx) [e^3] - (d/dx) [2^(3x-1)]
= e^(-√x) / (2x) + 0 - 3(2^(3x-1))ln2
This is the differentiation of the expression Y = e^(-√x) + e^3 - 2^(3x-1), without simplifying further.