Solution
9 (a) Using product rule
[tex]f(x)=(x-1)(3x+4)[/tex][tex]\begin{gathered} u=x-1 \\ \frac{du}{dx}=1 \end{gathered}[/tex][tex]\begin{gathered} v=3x+4 \\ \frac{dv}{dx}=3 \end{gathered}[/tex]
The product rule formula is quoted below
[tex]\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}[/tex][tex]\begin{gathered} f^{\prime}(x)=\frac{dy}{dx}=(x-1)\times3+(3x+4)\times1 \\ \\ f^{\prime}(x)=3(x-1)+3x+4 \\ f^{\prime}(x)=3x-3+3x+4 \\ f^{\prime}(x)=6x+1 \end{gathered}[/tex]
9(b) Verifying the resuts of 9(a) by bexpanding the product first and subsequently differentiating
[tex]\begin{gathered} f(x)=(x-1)(3x+4) \\ f(x)=3x^2+x-4 \\ \\ f^{\prime}(x)=\frac{dy}{dx}=6x+1 \end{gathered}[/tex]
In conclusion , we observe that the results of the derivatives are the same irrespective of the approach used as shown in 9(a) and 9(b) above
