c. find the coordinates of the stationary points on the curve ( y = x^{3}-6x^{2} ).\nd. find the second…

c. find the coordinates of the stationary points on the curve ( y = x^{3}-6x^{2} ).\nd. find the second derivative of ( y=(2x - 3)^{5} ). use chain rule\ne. find the equation of the normal to the curve ( y^{2}-x^{2}=5 ) at point ( (2,3) ).\nhint: use implicit differentiation\nf. differentiate ( f(x)=x - 2x^{2} ) using the definition ( f(x)=lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h} ).

c. find the coordinates of the stationary points on the curve ( y = x^{3}-6x^{2} ).\nd. find the second derivative of ( y=(2x - 3)^{5} ). use chain rule\ne. find the equation of the normal to the curve ( y^{2}-x^{2}=5 ) at point ( (2,3) ).\nhint: use implicit differentiation\nf. differentiate ( f(x)=x - 2x^{2} ) using the definition ( f(x)=lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h} ).

Answer

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}).