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Help needed on these questions
- Thread starter kalyan601
- Start date
HallsofIvy
Elite Member
- Joined
- Jan 27, 2012
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Yes, that is correct. As far as "format" is concerned, you might want to put the "-0.5" in parentheses.1. Given that y= 4x^3 - 1 +2x^0.5, x>0, find dy/dx
I think this is the answer but is this the correct way to format it?: dy/dx= 12x^2 +x^-0.5
No, this is not correct. Perhaps a couple of typos- I assume you mean "0.5", not "0/5"- but the derivative of -2x^2 is -4x, not "4".2. Given that y=4x + 3x^1.5 - 2x^2, x>0, find dy/dx
A: 4 + 4.5x^0/5 + 4
Are these correct?
Yeah I did mean 0.5. Thank you for your help. Stupid mistake by meNo, this is not correct. Perhaps a couple of typos- I assume you mean "0.5", not "0/5"- but the derivative of -2x^2 is -4x, not "4".
dy/dx = 12x^2 + x^-0.5
\(\displaystyle \text{Another way is } \dfrac{dy}{dx} = 12x^{2} + \dfrac{1}{\sqrt{x}}\)