What is the derivative of 4^[sin(pi*x)]? And how do you go about solving it? thanks
4^[sin(pi*x)] =
exp[log(4) sin(pi x)]
So, by the chain rule, the derivative is:
exp[log(4) sin(pi x)] log(4) pi
cos(pi x) =
pi log(4) 4^[sin(pi x)] cos(pi x)
To find the derivative of the function 4^[sin(pi*x)], we can use the chain rule.
First, let's rewrite the function in exponential form. 4^[sin(pi*x)] is equal to e^[ln(4^[sin(pi*x)])].
Next, rewrite the exponent using a logarithm property. ln(4^[sin(pi*x)]) is equal to (sin(pi*x)) * ln(4).
Now, we can apply the chain rule. The chain rule states that if we have a composite function f(g(x)), the derivative is given by f'(g(x)) * g'(x).
In our case, f(u) = e^u, and g(x) = (sin(pi*x)) * ln(4). The derivative of f(u) with respect to u is e^u, which means f'(g(x)) = e^((sin(pi*x)) * ln(4)).
To find g'(x), we need to differentiate (sin(pi*x)) * ln(4). The derivative of sin(pi*x) with respect to x is cos(pi*x) * pi, and the derivative of ln(4) with respect to x is 0 since ln(4) is a constant.
Therefore, g'(x) = (sin(pi*x)) * pi * 0 + cos(pi*x) * pi = pi * cos(pi*x).
Combining f'(g(x)) and g'(x) using the chain rule, we get the derivative of the original function:
Derivative of 4^[sin(pi*x)] = e^((sin(pi*x)) * ln(4)) * pi * cos(pi*x).
So, the derivative of the function 4^[sin(pi*x)] is e^((sin(pi*x)) * ln(4)) * pi * cos(pi*x).