Find the derivative of
a)y=3x^2Cos 2x
b)y=cot5x
c)y=2x^3+4x^2-3x+5
a) To find the derivative of y=3x^2Cos 2x, we will use the product rule.
Let u = 3x^2 and v = Cos 2x. Then, y = u * v.
Now, using the product rule:
y' = u'v + v'u
y' = (6x * Cos 2x) + (-2 * 3x^2 * Sin 2x)
y' = 6x * Cos 2x - 6x^2 * Sin 2x
Therefore, the derivative of y=3x^2Cos 2x is y' = 6x * Cos 2x - 6x^2 * Sin 2x.
b) To find the derivative of y=cot5x, we will use the chain rule.
Let u = 5x. Then, y = cot u = cot 5x.
Now, using the chain rule:
y' = -csc^2(5x) * 5
y' = -5csc^2(5x)
Therefore, the derivative of y=cot5x is y' = -5csc^2(5x).
c) To find the derivative of y=2x^3+4x^2-3x+5, we will differentiate each term individually.
y' = d(2x^3)/dx + d(4x^2)/dx - d(3x)/dx + d(5)/dx
y' = 6x^2 + 8x - 3
Therefore, the derivative of y=2x^3+4x^2-3x+5 is y' = 6x^2 + 8x - 3.