1. ## [SOLVED] Differentiation Question

Question:
The equation of a curve is $\displaystyle y = \frac{6}{5 - 2x}$
(i) Calculate the gradient of the curve at the point where $\displaystyle x = 1$
(ii) A point with coordinates $\displaystyle (x, y)$ moves along the curve in such a way that the rate of increase of $\displaystyle y$ has a constant value of $\displaystyle 0.02$ units per second. Find the rate of increase of $\displaystyle x$ when $\displaystyle x = 1$.

(i) $\displaystyle y = 6(5 - 2x)^{-1}$
$\displaystyle \frac{dy}{dx} = 6 \times -1(5 - 2x)^{-1-1} \times (-2)$
$\displaystyle \frac{dy}{dx} = 12(5 - 2x)^{-2}$

Is this right?

(ii) How can I get the rate of increase of $\displaystyle x$?

2. Hello ^^

Originally Posted by looi76
Question:
The equation of a curve is $\displaystyle y = \frac{6}{5 - 2x}$
(i) Calculate the gradient of the curve at the point where $\displaystyle x = 1$
(ii) A point with coordinates $\displaystyle (x, y)$ moves along the curve in such a way that the rate of increase of $\displaystyle y$ has a constant value of $\displaystyle 0.02$ units per second. Find the rate of increase of $\displaystyle x$ when $\displaystyle x = 1$.

(i) $\displaystyle y = 6(5 - 2x)^{-1}$
$\displaystyle \frac{dy}{dx} = 6 \times -1(5 - 2x)^{-1-1} \times (-2)$
$\displaystyle \frac{dy}{dx} = 12(5 - 2x)^{-2}$

Is this right?
Yes it is
(ii) How can I get the rate of increase of $\displaystyle x$?
The rate of increase is 0.02 means that $\displaystyle \frac{d^2y}{dx^2}=0.02$ (second derivative)