Determine derivative of each function
A) y = 3xSinx
B) f(t) = 2t^2Cos2t
C) [(pie)tSin](pie(t)-6)
To find the derivative of each function, we will apply the chain rule and product rule where necessary.
A) y = 3xSin(x)
To find the derivative, we need to apply the product rule. The product rule states that if we have two functions, u(x) and v(x), the derivative of their product is given by u'(x)v(x) + u(x)v'(x).
In this case, u(x) = 3x and v(x) = Sin(x). The derivative of u(x) is u'(x) = 3 and the derivative of v(x) is v'(x) = Cos(x).
Applying the product rule, we get:
y' = u'(x)v(x) + u(x)v'(x)
= (3)(Sin(x)) + (3x)(Cos(x))
= 3Sin(x) + 3xCos(x)
Therefore, the derivative of y = 3xSin(x) is y' = 3Sin(x) + 3xCos(x).
B) f(t) = 2t^2Cos(2t)
To find the derivative, we need to apply the product rule and the chain rule.
Let's break down the function into two functions: u(t) = 2t^2 and v(t) = Cos(2t).
The derivative of u(t) is u'(t) = 4t, and the derivative of v(t) is v'(t) = -2Sin(2t) by applying the chain rule.
Applying the product rule, we get:
f'(t) = u'(t)v(t) + u(t)v'(t)
= (4t)(Cos(2t)) + (2t^2)(-2Sin(2t))
= 4tCos(2t) - 4t^2Sin(2t)
Therefore, the derivative of f(t) = 2t^2Cos(2t) is f'(t) = 4tCos(2t) - 4t^2Sin(2t).
C) [(π)tSin][(π(t)-6)]
To find the derivative, we need to apply the product rule and the chain rule.
Let's break down the function into two functions: u(t) = (π)tSin(t) and v(t) = π(t - 6).
The derivative of u(t) involves applying the product rule: u'(t) = Sin(t) + tCos(t).
The derivative of v(t) is v'(t) = π by applying the chain rule.
Applying the product rule, we get:
f'(t) = u'(t)v(t) + u(t)v'(t)
= (Sin(t) + tCos(t))π(t - 6) + (π)tSin(t)
= π(t - 6)Sin(t) + tπ(t - 6)Cos(t) + πtSin(t)
Therefore, the derivative of [(π)tSin][(π(t)-6)] is f'(t) = π(t - 6)Sin(t) + tπ(t - 6)Cos(t) + πtSin(t).