Examples of General Theorems

Example:
Find \frac{{{\text{dy}}}}{{{\text{dx}}}} if {\text{y}}  = \left( {2{{\text{x}}^3} - 4{{\text{x}}^2}} \right)\left( {3{{\text{x}}^5} +  {{\text{x}}^2}} \right)
Solution:
We have

{\text{y}} = \left(  {2{{\text{x}}^3} - 4{{\text{x}}^2}} \right)\left( {3{{\text{x}}^5} +  {{\text{x}}^2}} \right)


Differentiate w.r.t ‘x’ and using product rule

\begin{gathered} \frac{{{\text{dy}}}}{{{\text{dx}}}} = \left( {2{{\text{x}}^3} -  4{{\text{x}}^2}} \right)\frac{{\text{d}}}{{{\text{dx}}}}\left( {3{{\text{x}}^5}  + {{\text{x}}^2}} \right) + \left( {3{{\text{x}}^5} + {{\text{x}}^2}}  \right)\frac{{\text{d}}}{{{\text{dx}}}}\left( {2{{\text{x}}^3} -  4{{\text{x}}^2}} \right) \\ \Rightarrow \frac{{{\text{dy}}}}{{{\text{dx}}}} = \left(  {2{{\text{x}}^3} - 4{{\text{x}}^2}} \right)\left(  {3\frac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^5} +  \frac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^2}} \right) + \left(  {3{{\text{x}}^5} + {{\text{x}}^2}} \right)\left(  {2\frac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^3} -  4\frac{{\text{d}}}{{{\text{dx}}}}{{\text{x}}^2}} \right) \\ \Rightarrow \frac{{{\text{dy}}}}{{{\text{dx}}}} = \left( {2{{\text{x}}^3}  - 4{{\text{x}}^2}} \right)\left( {3\left( {{\text{5}}{{\text{x}}^4}} \right) +  2{\text{x}}} \right) + \left( {3{{\text{x}}^5} + {{\text{x}}^2}} \right)\left(  {2\left( {{\text{3}}{{\text{x}}^2}} \right) - 4\left( {2{\text{x}}} \right)}  \right) \\ \Rightarrow \frac{{{\text{dy}}}}{{{\text{dx}}}} = \left(  {2{{\text{x}}^3} - 4{{\text{x}}^2}} \right)\left( {15{{\text{x}}^4} +  2{\text{x}}} \right) + \left( {3{{\text{x}}^5} + {{\text{x}}^2}} \right)\left(  {6{{\text{x}}^2} - 8{\text{x}}} \right) \\ \Rightarrow \frac{{{\text{dy}}}}{{{\text{dx}}}} = 30{{\text{x}}^7} +  4{{\text{x}}^4} - 60{{\text{x}}^6} - 8{{\text{x}}^3} + 18{{\text{x}}^7} -  24{{\text{x}}^6} + 6{{\text{x}}^4} - 8{{\text{x}}^3} \\ \Rightarrow \frac{{{\text{dy}}}}{{{\text{dx}}}} = 48{{\text{x}}^7} -  84{{\text{x}}^6} + 10{{\text{x}}^4} - 16{{\text{x}}^3} \\ \end{gathered}

Example:
Differentiate {\text{y}} = \frac{{\left( {2{{\text{x}}^3} + 4}  \right)}}{{\left( {{{\text{x}}^2} - 4{\text{x}} + 1} \right)}}with respect to ‘x’.
Solution:
We have

{\text{y}} = \frac{{\left(  {2{{\text{x}}^3} + 4} \right)}}{{\left( {{{\text{x}}^2} - 4{\text{x}} + 1}  \right)}}


Differentiate w.r.t ‘x’ and using quotient rule

\begin{gathered} \frac{{{\text{dy}}}}{{{\text{dx}}}} =  \frac{{\text{d}}}{{{\text{dx}}}}\frac{{\left( {2{{\text{x}}^3} + 4}  \right)}}{{\left( {{{\text{x}}^2} - 4{\text{x}} + 1} \right)}} \\ \Rightarrow  \frac{{{\text{dy}}}}{{{\text{dx}}}} = \frac{{\left( {{{\text{x}}^2} -  4{\text{x}} + 1} \right)\frac{{\text{d}}}{{{\text{dx}}}}\left( {2{{\text{x}}^3}  + 4} \right) - \left( {2{{\text{x}}^3} + 4}  \right)\frac{{\text{d}}}{{{\text{dx}}}}\left( {{{\text{x}}^2} - 4{\text{x}} +  1} \right)}}{{{{\left( {{{\text{x}}^2} - 4{\text{x}} + 1} \right)}^2}}} \\ \Rightarrow  \frac{{{\text{dy}}}}{{{\text{dx}}}} = \frac{{\left( {{{\text{x}}^2} - 4{\text{x}}  + 1} \right)\left( {{\text{6}}{{\text{x}}^2} + 0} \right) - \left(  {2{{\text{x}}^3} + 4} \right)\left( {{\text{2x}} - 4 + 0} \right)}}{{{{\left(  {{{\text{x}}^2} - 4{\text{x}} + 1} \right)}^2}}} \\ \Rightarrow \frac{{{\text{dy}}}}{{{\text{dx}}}}  = \frac{{\left( {{{\text{x}}^2} - 4{\text{x}} + 1} \right)\left(  {{\text{6}}{{\text{x}}^2}} \right) - \left( {2{{\text{x}}^3} + 4} \right)\left(  {{\text{2x}} - 4} \right)}}{{{{\left( {{{\text{x}}^2} - 4{\text{x}} + 1}  \right)}^2}}} \\ \Rightarrow  \frac{{{\text{dy}}}}{{{\text{dx}}}} = \frac{{6{{\text{x}}^4} - 24{{\text{x}}^3}  + 6{{\text{x}}^2} - \left( {4{{\text{x}}^4} - 8{{\text{x}}^3} + 8{\text{x}} -  16} \right)}}{{{{\left( {{{\text{x}}^2} - 4{\text{x}} + 1} \right)}^2}}} \\ \Rightarrow  \frac{{{\text{dy}}}}{{{\text{dx}}}} = \frac{{6{{\text{x}}^4} - 24{{\text{x}}^3}  + 6{{\text{x}}^2} - 4{{\text{x}}^4} + 8{{\text{x}}^3} - 8{\text{x}} +  16}}{{{{\left( {{{\text{x}}^2} - 4{\text{x}} + 1} \right)}^2}}} \\ \Rightarrow \frac{{{\text{dy}}}}{{{\text{dx}}}}  = \frac{{{\text{2}}{{\text{x}}^4} - 16{{\text{x}}^3} + 6{{\text{x}}^2} -  8{\text{x}} + 16}}{{{{\left( {{{\text{x}}^2} - 4{\text{x}} + 1} \right)}^2}}} \\ \end{gathered}