Answer :
Sure! Let's find the derivative of the given function step-by-step.
### Given:
[tex]\[ f(x) = 2x \sin x \cos x \][/tex]
### Step 1: Differentiate [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex].
To find the derivative [tex]\( f'(x) \)[/tex], we need to use the product rule and the chain rule. The product rule states that if you have a function [tex]\( g(x) = u(x)v(x) \)[/tex], then its derivative is given by:
[tex]\[ g'(x) = u'(x)v(x) + u(x)v'(x) \][/tex]
Let's set:
[tex]\[ u(x) = 2x \][/tex]
[tex]\[ v(x) = \sin x \cos x \][/tex]
First, we find the derivatives of [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex]:
[tex]\[ u'(x) = 2 \][/tex]
To find [tex]\( v'(x) \)[/tex], we use the product rule again since [tex]\( v(x) = \sin x \cos x \)[/tex]:
Let,
[tex]\[ v_1(x) = \sin x \][/tex]
and,
[tex]\[ v_2(x) = \cos x \][/tex]
Then,
[tex]\[ v'(x) = v_1'(x) v_2(x) + v_1(x) v_2'(x) \][/tex]
[tex]\[ v_1'(x) = \cos x \][/tex]
[tex]\[ v_2'(x) = -\sin x \][/tex]
Now substitute:
[tex]\[ v'(x) = (\cos x) (\cos x) + (\sin x) (-\sin x) \][/tex]
[tex]\[ v'(x) = \cos^2 x - \sin^2 x \][/tex]
Now we can find [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = u'(x) v(x) + u(x) v'(x) \][/tex]
[tex]\[ f'(x) = 2(\sin x \cos x) + 2x(\cos^2 x - \sin^2 x) \][/tex]
Simplify it:
[tex]\[ f'(x) = 2 \sin x \cos x + 2 x (\cos^2 x - \sin^2 x) \][/tex]
You can also write it as:
[tex]\[ f'(x) = -2x \sin^2 x + 2x \cos^2 x + 2 \sin x \cos x \][/tex]
So the derivative of the function is:
[tex]\[ f'(x) = -2x\sin^2 x + 2x\cos^2 x + 2\sin x\cos x \][/tex]
---
### Step 2: Evaluate [tex]\( f'(x) \)[/tex] at [tex]\( x = 4 \)[/tex].
To find [tex]\( f'(4) \)[/tex], substitute [tex]\( x = 4 \)[/tex] into [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(4) = -2(4)\sin^2(4) + 2(4)\cos^2(4) + 2\sin(4)\cos(4) \][/tex]
[tex]\[ f'(4) = -8\sin^2(4) + 8\cos^2(4) + 2\sin(4)\cos(4) \][/tex]
So the value of the derivative at [tex]\( x = 4 \)[/tex] is:
[tex]\[ f'(4) = -8\sin^2(4) + 2\sin(4)\cos(4) + 8\cos^2(4) \][/tex]
To summarize:
### Final Answers:
[tex]\[ f'(x) = -2x\sin^2 x + 2x\cos^2 x + 2\sin x\cos x \][/tex]
[tex]\[ f'(4) = -8\sin^2(4) + 2\sin(4)\cos(4) + 8\cos^2(4) \][/tex]
### Given:
[tex]\[ f(x) = 2x \sin x \cos x \][/tex]
### Step 1: Differentiate [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex].
To find the derivative [tex]\( f'(x) \)[/tex], we need to use the product rule and the chain rule. The product rule states that if you have a function [tex]\( g(x) = u(x)v(x) \)[/tex], then its derivative is given by:
[tex]\[ g'(x) = u'(x)v(x) + u(x)v'(x) \][/tex]
Let's set:
[tex]\[ u(x) = 2x \][/tex]
[tex]\[ v(x) = \sin x \cos x \][/tex]
First, we find the derivatives of [tex]\( u(x) \)[/tex] and [tex]\( v(x) \)[/tex]:
[tex]\[ u'(x) = 2 \][/tex]
To find [tex]\( v'(x) \)[/tex], we use the product rule again since [tex]\( v(x) = \sin x \cos x \)[/tex]:
Let,
[tex]\[ v_1(x) = \sin x \][/tex]
and,
[tex]\[ v_2(x) = \cos x \][/tex]
Then,
[tex]\[ v'(x) = v_1'(x) v_2(x) + v_1(x) v_2'(x) \][/tex]
[tex]\[ v_1'(x) = \cos x \][/tex]
[tex]\[ v_2'(x) = -\sin x \][/tex]
Now substitute:
[tex]\[ v'(x) = (\cos x) (\cos x) + (\sin x) (-\sin x) \][/tex]
[tex]\[ v'(x) = \cos^2 x - \sin^2 x \][/tex]
Now we can find [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(x) = u'(x) v(x) + u(x) v'(x) \][/tex]
[tex]\[ f'(x) = 2(\sin x \cos x) + 2x(\cos^2 x - \sin^2 x) \][/tex]
Simplify it:
[tex]\[ f'(x) = 2 \sin x \cos x + 2 x (\cos^2 x - \sin^2 x) \][/tex]
You can also write it as:
[tex]\[ f'(x) = -2x \sin^2 x + 2x \cos^2 x + 2 \sin x \cos x \][/tex]
So the derivative of the function is:
[tex]\[ f'(x) = -2x\sin^2 x + 2x\cos^2 x + 2\sin x\cos x \][/tex]
---
### Step 2: Evaluate [tex]\( f'(x) \)[/tex] at [tex]\( x = 4 \)[/tex].
To find [tex]\( f'(4) \)[/tex], substitute [tex]\( x = 4 \)[/tex] into [tex]\( f'(x) \)[/tex]:
[tex]\[ f'(4) = -2(4)\sin^2(4) + 2(4)\cos^2(4) + 2\sin(4)\cos(4) \][/tex]
[tex]\[ f'(4) = -8\sin^2(4) + 8\cos^2(4) + 2\sin(4)\cos(4) \][/tex]
So the value of the derivative at [tex]\( x = 4 \)[/tex] is:
[tex]\[ f'(4) = -8\sin^2(4) + 2\sin(4)\cos(4) + 8\cos^2(4) \][/tex]
To summarize:
### Final Answers:
[tex]\[ f'(x) = -2x\sin^2 x + 2x\cos^2 x + 2\sin x\cos x \][/tex]
[tex]\[ f'(4) = -8\sin^2(4) + 2\sin(4)\cos(4) + 8\cos^2(4) \][/tex]
