Long Questions (Question 5)

Question 5 (7 marks):
It is given that the equation of a curve is $y=\frac{5}{x^2}.$  
(a) Find the value of $\frac{dy}{dx}$ when x = 3.
(b) Hence, estimate the value of $\frac{5}{{\left(2.98\right)}^2}.$  

Solution:
(a)
$\begin{array}{l}y=\frac{5}{x^2}=5x^{-2}\\ \frac{dy}{dx}=-10x^{-3}\\ =-\frac{10}{x^3}\\ \text{When }x=3\\ \frac{dy}{dx}=-\frac{10}{3^3}=-\frac{10}{27}\end{array}$


(b)
$\begin{array}{l}\delta x=2.98-3=-0.02\\ \delta y=\frac{dy}{dx}.\delta x\\ =-\frac{10}{27}\times \left(-0.02\right)\\ =0.007407\\ \\ \text{Values of }\frac{5}{{\left(2.98\right)}^2}\\ =y+\delta y\\ =\frac{5}{x^2}+\left(0.007407\right)\\ =\frac{5}{3^2}+\left(0.007407\right)\\ =0.56296\end{array}$