What is the derivative of 7e^(3sin^(4x+5))?
typo ...
3sin^(4x+5) has no argument
looks like 4x+5 is the exponent, but there is no "angle"
To find the derivative of the function 7e^(3sin^(4x+5)), we can use the chain rule.
The chain rule states that if we have a composition of functions, such as f(g(x)), the derivative of f(g(x)) with respect to x is given by f'(g(x)) multiplied by g'(x).
In this case, let's break down the function:
f(x) = 7e^x
g(x) = 3sin^(4x+5)
Now we can find the derivatives of f(x) and g(x):
f'(x) = 7e^x (the derivative of e^x is e^x)
g'(x) = 12cos(4x+5) (the derivative of sin^x is cos(x) and the derivative of 4x+5 is 4)
Next, we substitute these derivatives into the chain rule:
(f(g(x)))' = f'(g(x)) * g'(x)
= 7e^(3sin^(4x+5)) * 12cos(4x+5)
Therefore, the derivative of 7e^(3sin^(4x+5)) is 84e^(3sin^(4x+5))cos(4x+5).