the derivative of 4xsin(x)cos(x)
using the product rule,
d/dx (4x)(sinx cosx)
= (4)(sinx cosx) + (4x) d/dx (sinx)(cosx)
= 4sinx cosx + 4x(cos^2x - sin^2x)
= 4(sinx cosx + x (cos^2x - sin^2x))
or, for less work,
4x sinx cosx = 2x sin2x
so the derivative is
2 sin2x + 4x cos2x
To find the derivative of the function f(x) = 4xsin(x)cos(x), we can use the product rule and the chain rule of differentiation.
Step 1: Apply the product rule.
The product rule states that for two functions u(x) and v(x), the derivative of their product is given by:
(d/dx)(u(x) * v(x)) = u'(x) * v(x) + u(x) * v'(x)
In this case, u(x) = 4x and v(x) = sin(x)cos(x).
Step 2: Compute the derivatives of u(x) and v(x).
The derivative of u(x) = 4x with respect to x is u'(x) = 4.
To find the derivative of v(x), we can use the chain rule. The chain rule states that for a composition of functions h(g(x)), the derivative is given by:
(d/dx)(h(g(x))) = h'(g(x)) * g'(x)
In this case, h(x) = sin(x)cos(x).
The derivative of h(x) with respect to x is h'(x) = cos(x)cos(x) - sin(x)sin(x).
Using the trigonometric identity cos(2x) = cos^2(x) - sin^2(x), we can simplify h'(x) to cos(2x).
Step 3: Apply the product and chain rule to find the derivative.
Using the product rule and the derivatives we computed in Step 2, we have:
f'(x) = u'(x) * v(x) + u(x) * v'(x)
= 4 * (sin(x)cos(x)) + 4x * cos(2x)
Therefore, the derivative of f(x) = 4xsin(x)cos(x) is f'(x) = 4sin(x)cos(x) + 4xcos(2x).
To find the derivative of the function 4xsin(x)cos(x), we can use the product rule and the chain rule.
Step 1: Apply the product rule.
The product rule states that the derivative of the product of two functions, u(x) and v(x), is given by the formula:
(uv)' = u'v + uv'
Let's define u(x) = 4x and v(x) = sin(x)cos(x).
Then, u'(x) = 4 and v'(x) = (sin(x)cos(x))'.
Step 2: Find the derivative of v(x) using the chain rule.
The chain rule states that if y = f(g(x)), then the derivative of y with respect to x is given by the formula:
dy/dx = f'(g(x)) * g'(x)
Let's define f(u) = sin(u) and g(x) = cos(x).
Then, f'(u) = cos(u) and g'(x) = -sin(x).
Now, we can find (sin(x)cos(x))' using the chain rule as follows:
(sin(x)cos(x)) = f(g(x)) = f(cos(x)) = sin(cos(x))
Therefore, (sin(x)cos(x))' = f'(g(x)) * g'(x) = cos(cos(x)) * (-sin(x))
Step 3: Apply the product rule to get the derivative of 4xsin(x)cos(x).
Using the product rule formula mentioned earlier, we have:
(4xsin(x)cos(x))' = (4x)'(sin(x)cos(x)) + (4x)(sin(x)cos(x))'
Substituting the values we obtained earlier, we get:
= 4(sin(x)cos(x)) + 4x(cos(cos(x)) * (-sin(x)))
Simplifying further, we have:
= 4sin(x)cos(x) - 4xsin(x)cos(cos(x))
Thus, the derivative of 4xsin(x)cos(x) is 4sin(x)cos(x) - 4xsin(x)cos(cos(x)).