d. find the second derivative of ( y=(2 x - 3)^{5} ). use chain rule\ne. find the equation of the normal to…

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}).

Answer:

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

Explanation 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 for part e:

(y =-\frac{3}{2}x + 6)