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?

Question

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?

myininaya · Answer

$$y=\sqrt{x}(3x-1)$$?

anonymous · Answer

yes

myininaya · Answer

Do you want to do the product rule and power rule?
Or!
Do you want to distribute and then do derivative junk?

anonymous · Answer

9x-1/2sqrtx is right

anonymous · Answer

i tried using the product and power rules.

myininaya · Answer

K 
$$(\sqrt{x})' =? , (3x-1)'=?$$

anonymous · Answer

1/2(x)^-1/2 , 3?

anonymous · Answer

$$\sqrt{x}(3x-1)=3x^{\frac{ 3 }{ 2 }}-x^{\frac{ 1 }{ 2 }}$$$$\frac{ d }{ dx }(3x^{\frac{ 3 }{ 2 }}-x^{\frac{ 1 }{ 2 }})=\frac{ 9 }{ 2 }x^{\frac{ 1 }{ 2 }}-\frac{ 1 }{ 2 }x^{\frac{ -1 }{ 2 }}$$

anonymous · Answer

Now rearrange the expression above so as to get what you miss.