evaluate the derivative of f(X)= 3Xsin(X) at a=0
is that at a=0 or x=0,,
anyway, to get the derivative, do the chain rule since two different terms with x are multiplied,, to do this let g(x)=3x and h(x)=sin x
thus d/dx [g(x)h(x)] = g(x)*h'(x) + g'(x)h(x)
*g(x)*h'(x) = you get derivative of h(x)=sin x an multiply this to g(x)=3x
*g'(x)h(x) = you get derivative of g(x)=3x an multiply this to h(x)=sin x
so there,, please ask questions if you did not understand~
I'm not too sure if it's a=0 or x=0... becuase my worksheet said a=0. Anyways after i use the rule, would I need to plug in the zero?
To evaluate the derivative of f(x) = 3xsin(x) at a = 0, we need to use the product rule and the chain rule.
1. Apply the product rule: The derivative of a product of two functions u(x) and v(x) is given by (u(x)v'(x) + v(x)u'(x)).
In this case, let u(x) = 3x and v(x) = sin(x).
2. Find the derivative of u(x) = 3x.
The derivative of u(x) with respect to x is simply 3.
3. Find the derivative of v(x) = sin(x).
The derivative of sin(x) with respect to x is cos(x).
4. Apply the product rule to find the derivative of f(x).
f'(x) = u(x)v'(x) + v(x)u'(x)
= (3x)(cos(x)) + (sin(x))(3)
= 3xcos(x) + 3sin(x)
5. Evaluate the derivative at a = 0.
Substitute x = 0 into the derived equation:
f'(0) = 3(0)cos(0) + 3sin(0)
= 0 + 0
= 0
Therefore, the derivative of f(x) = 3xsin(x) at a = 0 is 0.