Find the gradient of the curve $y=\frac{2x-4}{x^2}$ at the point where the curve crosses the axis.
I did it
$y=\frac{2}{2x}$
but the right answer is $\frac{1}{2}$
Find the gradient of the curve $y=\frac{2x-4}{x^2}$ at the point where the curve crosses the axis.
I did it
$y=\frac{2}{2x}$
but the right answer is $\frac{1}{2}$
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:
y'=\frac{2x^2-2x(2x-4)}{x^4}\tag{1}
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}$ $= \frac{-2x+8}{x^3}\tag{2}$
$(2)\stackrel{\text{x=2}}{\implies}\frac{-2 \cdot 2+8}{2^3}=\frac{1}{2}$
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 $\nabla y|_{x=2}=\frac{-2x+8}{x^3}\vec{i}|_{x=2}=\frac{1}{2}\vec{i}$
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} $