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Find dy/dx for y=(sqrt x)(3x-1). I got the answer 9x-3/2sqrtx, but the answer is supposed to be 9x-1/2sqrtx. Can anybody help me understand what I did wrong?
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\[y=\sqrt{x}(3x-1)\]?
yes
Do you want to do the product rule and power rule? Or! Do you want to distribute and then do derivative junk?
9x-1/2sqrtx is right
i tried using the product and power rules.
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K \[(\sqrt{x})' =? , (3x-1)'=?\]
1/2(x)^-1/2 , 3?
\[\sqrt{x}(3x-1)=3x^{\frac{ 3 }{ 2 }}-x^{\frac{ 1 }{ 2 }}\]\[\frac{ d }{ dx }(3x^{\frac{ 3 }{ 2 }}-x^{\frac{ 1 }{ 2 }})=\frac{ 9 }{ 2 }x^{\frac{ 1 }{ 2 }}-\frac{ 1 }{ 2 }x^{\frac{ -1 }{ 2 }}\]
Now rearrange the expression above so as to get what you miss.
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