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Find the gradient of the curve $y=\frac{2x-4}{x^2}$ at the point where the curve crosses the axis.

I did it
but the right answer is $\frac{1}{2}$

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up vote 3 down vote accepted

The point where the curve crosses the axis is $(2,0)$.

To find the gradient, you need to find the first derivative of the function:


And all you should do is calculating $y'$ when $x=2$:

$$(1)\stackrel{\text{(2,y')}}{\implies} y'=\frac{1}{2}$$

Your derivative is wrong, Using the quotient rule, you'll have:

$$y' = \frac{(2x-4)' \cdot x^2 - (x^2)' \cdot (2x-4)}{{(x^2)}^2}$$

$$ = \frac{2 \cdot x^2 - 2x \cdot (2x-4)}{x^4}$$

$$ = \frac{-2x^2 +8x }{x^4}$$ $$= \frac{-2x+8}{x^3}\tag{2}$$

$$(2)\stackrel{\text{x=2}}{\implies}\frac{-2 \cdot 2+8}{2^3}=\frac{1}{2}$$

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why not derivative is $y=\frac{2}{2x}$ and how did it into $y'=\frac{2x^2-2x(2x-4)}{x^4}$ can u explain? thx – Sb Sangpi Apr 12 '12 at 13:19
@SbSangpi: Did you use the quotient rule? You don't need to simplify it since you will set $x=2$. It's wrong, try again using the rule and you'l go fine. – Gigili Apr 12 '12 at 13:34
$\frac{-2x+8}{x^3}$ shouldn't be $\frac{-2x+8}{4x^3}$ or u cancel by $x$? thx alot. – Sb Sangpi Apr 12 '12 at 23:20
@SbSangpi: Yes, an $x$ in the numerator with an $x$ in the denominator. – Gigili Apr 13 '12 at 18:01

Note that, in general, $$ \left(\frac{u}{v}\right)'\not=\frac{u'}{v'} $$ When taking the derivative of a quotient, the rule is $$ \left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2} $$ which, as pointed out in other posts, is $$ \frac{2\cdot x^2-(2x-4)\cdot2x}{x^4}=\frac{8-2x}{x^3} $$

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The gradient at a specific point is a fixed vector, while the gradient function is a function of the independent variables. So therefore you have to compute the gradient at the specific point. In this case the point where the curve crosses the axis, so (2,0). So therefore $\nabla y|_{x=2}=\frac{-2x+8}{x^3}\vec{i}|_{x=2}=\frac{1}{2}\vec{i}$

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