differentiate y = 2^3x^2
read as two to the 3x squared power
y=a^kx
y'= k*a^(kx)*ln(a)
y = 2^(3x^2)
y' = ln2 * 6x * 2^(3x^2)
it helps to go at it like this
y = 2^(3x^2)
ln y = 3x^2 ln2
1/y y' = 6x ln2
y' = ln2 6x 2^(3x^2)
To differentiate the function y = 2^(3x^2), we can use the chain rule of differentiation. The chain rule states that if we have a function of the form f(g(x)), where f(u) and g(x) are both differentiable functions, then the derivative of f(g(x)) is given by f'(g(x)) * g'(x).
In this case, let's consider f(u) = 2^u and g(x) = 3x^2. We want to find the derivative of f(g(x)) = 2^(3x^2).
Step 1: Find the derivative of f(u) = 2^u.
The derivative of 2^u with respect to u is given by ln(2) * 2^u. (This is a standard result based on logarithmic differentiation.)
Step 2: Find the derivative of g(x) = 3x^2.
The derivative of 3x^2 with respect to x is 6x.
Step 3: Apply the chain rule.
The derivative of f(g(x)) = 2^(3x^2) is given by f'(g(x)) * g'(x), which becomes ln(2) * 2^(3x^2) * 6x.
Therefore, the derivative of y = 2^(3x^2) is dy/dx = ln(2) * 2^(3x^2) * 6x.