Question

Find the gradient of the curve y=(x-2)/square root x at the point where x=4. I am close to the answer but not quite there, please help!

Answer 1

Heres the curve again:\[y=\frac{ x-2 }{ \sqrt{x} }\]

Answer 2

I tried differentiating and got this:\[-\frac{ 1 }{ 2 } x \times x ^{\frac{ -3 }{ 2 }}+ x \frac{ -3 }{ 2 }\]

Answer 3

But then when I plug in x I get \[-\frac{ 1 }{ 8 }\]

Answer 4

And according to textbook thats not right!

Answer 5

you haven't differentiated correctly...what answer does the text give?

Answer 6

The text book gives the answer as 3/8

Answer 7

\[y=\frac{ x-2 }{ \sqrt{x} } = x^{1/2} - 2x^{-1/2}\]
\[y' = {1 \over 2} x^{-1/2} + x^{-3/2}\]

Answer 8

so \[\frac{ x }{ \sqrt{x} }= x ^{\frac{ 1 }{ 2 }}\]

Answer 9

\[\frac{ x^1 }{ \sqrt{x} }= x ^{(1-\frac{ 1 }{ 2 })} = x^{1/2}\]

Answer 10

oh I see. Ok I will try again.

Answer 11

Got the right answer now, thank you very much!