Find y' if y = 2^(-x^2 )

To find y', we need to use the chain rule. The chain rule states that if we have a function of the form f(g(x)), then the derivative of f(g(x)) with respect to x is given by f'(g(x)) * g'(x).

In this case, our function is y = 2^(-x^2). Let's break it down:

Step 1: Find the derivative of the outer function f(g(x)). The outer function is f(u) = 2^u, where u = -x^2. To find the derivative of f(u), we use the power rule, which states that the derivative of a constant raised to the power of u is given by the constant multiplied by the natural logarithm of the base, multiplied by u. In this case, the base is 2, so the derivative of f(u) = 2^u is f'(u) = (2^u) * ln(2).

Step 2: Find the derivative of the inner function g(x). The inner function is g(x) = -x^2. To find the derivative of g(x), we use the power rule again. The derivative of g(x) = -x^2 is g'(x) = -2x.

Step 3: Apply the chain rule. Now that we have the derivatives of both the outer and inner functions, we can apply the chain rule. The derivative of y = 2^(-x^2) with respect to x is given by y' = f'(g(x)) * g'(x). Substituting the derivatives we found in steps 1 and 2, we have y' = (2^(-x^2)) * ln(2) * (-2x).

So, the derivative of y = 2^(-x^2) is y' = (2^(-x^2)) * ln(2) * (-2x).