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 $\frac{1}{2}$
[Math] Find the gradient of the curve
calculusderivatives
calculusderivatives
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 $\frac{1}{2}$
Best Answer
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:
$$y' = \frac{(2x-4)' \cdot x^2 - (x^2)' \cdot (2x-4)}{{(x^2)}^2}$$
$$ = \frac{2 \cdot x^2 - 2x \cdot (2x-4)}{x^4}$$
$$ = \frac{-2x^2 +8x }{x^4}$$ $$= \frac{-2x+8}{x^3}\tag{2}$$
$$(2)\stackrel{\text{x=2}}{\implies}\frac{-2 \cdot 2+8}{2^3}=\frac{1}{2}$$