Question

c. find the coordinates of the stationary points on the curve ( y = x^{3}-6x^{2} ).
d. find the second derivative of ( y=(2x - 3)^{5} ). use chain rule
e. find the equation of the normal to the curve ( y^{2}-x^{2}=5 ) at point ( (2,3) ).
hint: use implicit differentiation
f. differentiate ( f(x)=x - 2x^{2} ) using the definition ( f(x)=lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h} ).

Explanation:

Step1: Find the first derivative

Given \( y=(2x - 3)^5 \). Let \( u = 2x-3 \), then \( y = u^5 \).
By the chain - rule \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \).
\(\frac{dy}{du}=5u^{4}\) and \(\frac{du}{dx}=2\).
So \(\frac{dy}{dx}=5(2x - 3)^{4}\cdot2=10(2x - 3)^{4}\).

Step2: Find the second derivative

Let \( v=(2x - 3)^{4} \), and \( z = 10v \).
Again, by the chain - rule \(\frac{dz}{dx}=\frac{dz}{dv}\cdot\frac{dv}{dx}\).
\(\frac{dz}{dv}=10\).
For \(v=(2x - 3)^{4}\), let \(t = 2x-3\), then \(v=t^{4}\).
\(\frac{dv}{dt}=4t^{3}\) and \(\frac{dt}{dx}=2\). So \(\frac{dv}{dx}=4(2x - 3)^{3}\cdot2 = 8(2x - 3)^{3}\).
Then \(\frac{d^{2}y}{dx^{2}}=\frac{dz}{dx}=10\times8(2x - 3)^{3}=80(2x - 3)^{3}\).

Answer:

The second derivative of \(y=(2x - 3)^5\) is \(80(2x - 3)^{3}\).