What is the slope of the tangent of each curve at given point
y=√(16x^3), (4,32)
If I find the derivative of y=√(16x^3)
Will it be 16/-x^3
nope. √x^3 = x^(3/2)
so,
y = √16x^3 = 4x^(3/2)
y' = 4(3/2)x^(1/2) = 6√x
y'(4) = 6√4 = 12
y=√(16x^3), (4,32)
If I find the derivative of y=√(16x^3)
Will it be 16/-x^3
so,
y = √16x^3 = 4x^(3/2)
y' = 4(3/2)x^(1/2) = 6√x
y'(4) = 6√4 = 12