Question

d. find the second derivative of ( y=(2 x - 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

Explanation:

Step1: Find the first derivative

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 \(v=(2x - 3)^{4}\), then \(\frac{dy}{dx}=10v\).
By the chain rule \(\frac{d^{2}y}{dx^{2}}=10\frac{dv}{dx}\).
Let \(t = 2x - 3\), \(v=t^{4}\), \(\frac{dv}{dt}=4t^{3}\), \(\frac{dt}{dx}=2\).
So \(\frac{dv}{dx}=4(2x - 3)^{3}\cdot2=8(2x - 3)^{3}\).
Then \(\frac{d^{2}y}{dx^{2}}=10\times8(2x - 3)^{3}=80(2x - 3)^{3}\).

for part e:

Step1: Differentiate implicitly

Differentiate \(y^{2}-x^{2}=5\) with respect to \(x\).
Using the chain - rule for \(y^{2}\) (since \(y\) is a function of \(x\)): \(2y\frac{dy}{dx}-2x = 0\).
Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx}=\frac{x}{y}\).

Step2: Find the slope of the tangent at \((2,3)\)

Substitute \(x = 2\) and \(y = 3\) into \(\frac{dy}{dx}\), \(\frac{dy}{dx}\big|_{(2,3)}=\frac{2}{3}\).

Step3: Find the slope of the normal

The slope of the normal \(m_{n}\) is the negative reciprocal of the slope of the tangent. So \(m_{n}=-\frac{3}{2}\).

Step4: Use the point - slope form

The point - slope form of a line is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})=(2,3)\) and \(m =-\frac{3}{2}\).
\(y - 3=-\frac{3}{2}(x - 2)\).
Expand: \(y-3=-\frac{3}{2}x + 3\).
\(y=-\frac{3}{2}x+6\).

Answer:

\(y'' = 80(2x - 3)^{3}\)