Differentiate:
F(x)=e^(xsin2x)
This isn't homework, it's review for our exam. Couldn't remember how to work it :( Thanks!
Several layers of derivatives here ...
remember for y = e^u
dy/dx = e^u (du/dx)
so f'(x) = e^(xsin2x)(d(xsin2x)/dx)
= e^(xsin2x)(x(2cos2x) + sin2x) using the product rule for the last part
Thank you! :)
To differentiate the function F(x) = e^(xsin2x), we will use the chain rule. The chain rule states that if we have a composite function, where one function is inside another, we can differentiate it by differentiating the outer function and then multiplying it by the derivative of the inner function.
Let's break down the function F(x) = e^(xsin2x) into its composite functions:
1. Outer function: f(u) = e^u, where u = xsin2x
2. Inner function: u = xsin2x
Now, let's differentiate each function separately:
Derivative of the inner function (u = xsin2x):
To differentiate u with respect to x, we will apply the product rule.
Let v = x and w = sin2x.
Using the product rule, we have:
du/dx = v(d/dx(w)) + w(d/dx(v))
=> du/dx = 1 * (d/dx(sin2x)) + sin2x * (d/dx(x))
=> du/dx = 2cos2x + xcos2x
Derivative of the outer function (f(u) = e^u):
To differentiate f(u) with respect to u, we simply take the derivative of e^u, which is e^u itself.
Now, we can apply the chain rule by multiplying the derivatives together:
d/dx[F(x)] = d/du[f(u)] * du/dx
= e^u * (2cos2x + xcos2x)
= e^(xsin2x) * (2cos2x + xcos2x)
Therefore, the derivative of F(x) = e^(xsin2x) is d/dx[F(x)] = e^(xsin2x) * (2cos2x + xcos2x).