Asked by Confused
Also I would like to know how to set this problem up in finding the answer:
Find the second derivative.
y= sqrt(3x+4)
Thanks, P.S. I tried taking 3x+4 and squaring it to the 1/2. I thought that might've been the same, then I took the derivative of that and got 1/2(3x+4)^(-1/2), then took the 2nd derivative of that and got -1/4(3x+4)^(-1.5). Apparently this isn't right. What am I doing wrong?
y=(3x+4)^(1/2)
y'=(3/2)(3x+4)^(-1/2)
y"=(-9/4)(3x+4)^(-3/2)
i think this is correct. You forgot the chain rule. You also have to multiply by the derivative of 3x+4
THANK U!!
Find the second derivative.
y= sqrt(3x+4)
Thanks, P.S. I tried taking 3x+4 and squaring it to the 1/2. I thought that might've been the same, then I took the derivative of that and got 1/2(3x+4)^(-1/2), then took the 2nd derivative of that and got -1/4(3x+4)^(-1.5). Apparently this isn't right. What am I doing wrong?
y=(3x+4)^(1/2)
y'=(3/2)(3x+4)^(-1/2)
y"=(-9/4)(3x+4)^(-3/2)
i think this is correct. You forgot the chain rule. You also have to multiply by the derivative of 3x+4
THANK U!!
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