Find the derivative for k(x) = (5x4 + 2)(3sin x)
f(uv)=v(f'(u) + u(f'(v))
(5x^4+2)(-3cosx) + 3sinx(4x^3)
Seems to me that
k(x) = (5x^4 + 2)(3sin x)
k' = (20x^3)(3sinx) + (5x^4+2)(3cosx)
= 15x^3(4sinx + xcosx) + 6cosx
To find the derivative of the given function k(x) = (5x^4 + 2)(3sin x), we will use the product rule of differentiation. The product rule states that if we have two functions, f(x) and g(x), the derivative of their product is given by:
(f(x) * g(x))' = f'(x) * g(x) + f(x) * g'(x)
So, let's apply the product rule to find the derivative of k(x):
First, let's find the derivative of the first function, f(x) = (5x^4 + 2):
f'(x) = d/dx (5x^4 + 2)
= 20x^3
Now, let's find the derivative of the second function, g(x) = 3sin(x):
g'(x) = d/dx (3sin(x))
= 3cos(x)
Finally, we can apply the product rule formula to find the derivative of k(x):
k'(x) = f'(x) * g(x) + f(x) * g'(x)
= (20x^3) * (3sin(x)) + (5x^4 + 2) * (3cos(x))
Therefore, the derivative of k(x) = (5x^4 + 2)(3sin(x)) is:
k'(x) = (20x^3) * (3sin(x)) + (5x^4 + 2) * (3cos(x))