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