Homework Help: Calculus-derivatives

Posted by Jake on Wednesday, February 15, 2012 at 10:17pm.

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

Calculus-derivatives - Steve, Thursday, February 16, 2012 at 11:42am
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

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