If y = aeˣ+ be⁻ˣ + c Where a, b, c are parameters, they y’ is equal to

The correct option is ae^x – be^–x.

To find the first derivative of the given function, let us start with the equation:
$$y=a{e}^x+b{e}^{-x}+c$$
where _a, _b, and _c are constant parameters.

We differentiate both sides with respect to _x to find _y' (or _dy/dx):
$$y^{\prime }=\frac{d}{dx}(a{e}^x+b{e}^{-x}+c)$$

Using the sum rule of differentiation, we can differentiate each term individually:
$$y^{\prime }=\frac{d}{dx}\left(a{e}^x\right)+\frac{d}{dx}\left(b{e}^{-x}\right)+\frac{d}{dx}\left(c\right)$$

Now, let us evaluate the derivative of each term:
1. For the first term, _a is a constant coefficient, and the derivative of _e^x with respect to _x is _e^x:
$$\frac{d}{dx}\left(a{e}^x\right)=a{e}^x$$
2. For the second term, _b is a constant coefficient. Applying the chain rule to _e^–x, we get:
$$\frac{d}{dx}\left(b{e}^{-x}\right)=b\cdot (e^{-x}\cdot \frac{d}{dx}(-x))=b\cdot (-e^{-x})=-b{e}^{-x}$$
3. For the third term, _c is a constant parameter, and the derivative of any constant is 0:
$$\frac{d}{dx}\left(c\right)=0$$

Combining all the terms back together, we get:
$$y^{\prime }=a{e}^x-b{e}^{-x}$$