d/dx(e^sinx2x)=
a)-cos2xe^sinx2x
b)cosxe^sin2x
c)2e^sin2x
d)2cos2xe^sin2x
e)-2cos2xe^sin2x
use chain rule
d/dx(e^u) = e^u du/dx
where u = sin2x
have you any ideas now?
To find the derivative of the function f(x) = e^sin(x^2x) with respect to x, we can use the chain rule.
The chain rule states that if we have a composition of functions, f(g(x)), then the derivative of f(g(x)) with respect to x is given by f'(g(x)) * g'(x).
Let's break down the function f(x) = e^sin(x^2x) into the composition of functions:
g(x) = sin(x^2x)
f(u) = e^u, where u = g(x)
Now, let's find the derivative of the inner function g(x) using the chain rule:
g'(x) = d/dx(sin(x^2x))
To find this derivative, we can apply the chain rule again:
Let h(x) = x^2x, then
g'(x) = d/dx(sin(h(x))) = cos(h(x)) * h'(x)
Now, let's find h'(x):
h(x) = x^2x
Taking the derivative using the product rule:
h'(x) = (2x * x^2x-1) + (ln(x^2) * x^2x)
Now, let's substitute the values back into g'(x):
g'(x) = cos(x^2x) * [(2x * x^2x-1) + (ln(x^2) * x^2x)]
Finally, let's find the derivative of f(x) = e^sin(x^2x) using the chain rule:
f(u) = e^u
f'(u) = e^u
Now, substituting u = g(x) and g(x) = sin(x^2x):
f'(g(x)) = e^sin(x^2x)
Multiplying f'(g(x)) and g'(x):
f'(g(x)) * g'(x) = e^sin(x^2x) * cos(x^2x) * [(2x * x^2x-1) + (ln(x^2) * x^2x)]
Therefore, the correct answer is d) 2cos(2x)e^sin(2x).