To find the derivative of the function given we can use the product rule and the quotient rule.
Product rule:
The product rule states that:
[tex]\frac{d}{dx}(fg)=g\frac{df}{dx}+f\frac{dg}{dx}[/tex]Then, for the function given we have:
[tex]\begin{gathered} \frac{d}{ds}\lbrack(s^3+4)(4s^2+6)\rbrack=(4s^2+6)\frac{d}{ds}(s^3+4)+(s^3+4)\frac{d}{ds}(4s^2+6) \\ =(4s^2+6)(3s^2)+(s^3+4)(8s) \\ =12s^4+18s^2+8s^4+32s \\ =20s^4+18s^2+32s \end{gathered}[/tex]therefore the derivative is:
[tex]\frac{dg}{ds}=20s^4+18s^2+32s[/tex]Quotient rule:
To use the quotient rule we need to write the function in the following way:
[tex]g(s)=\frac{s^3+4}{(4s^2+6)^{-1}}[/tex]Now, the quotient rulre states that:
[tex]\frac{d}{dx}(\frac{f}{g})=\frac{g\frac{df}{dx}-f\frac{dg}{dx}}{g^2}[/tex]Then, in our case we have:
[tex]\begin{gathered} \frac{d}{ds}(\frac{s^3+4}{(4s^2+6)^{-1}})=\frac{(4s^2+6)^{-1}\frac{d}{ds}(s^3+4)-(s^3+4)\frac{d}{dx}(4s^2+6)^{-1}}{\lbrack(4s^2+6)^{-1}\rbrack^2} \\ =\frac{(4s^2+6)^{-1}(3s^2)-(s^3+4)\lbrack-(4s^2+6)^{-2}\rbrack^{}(8s)}{\lbrack(4s^2+6)^{-1}\rbrack^2} \\ =\frac{\frac{3s^2}{4s^2+6}+\frac{8s(s^3+4)}{(4s^2+6)^2}}{\lbrack(4s^2+6)^{-1}\rbrack^2} \\ =\frac{\frac{1}{4s^2+6}(3s^2+\frac{8s^4+32s}{4s^2+6})}{\lbrack(4s^2+6)^{-1}\rbrack^2} \\ =\frac{\frac{1}{4s^2+6}(\frac{3s^2(4s^2+6)+8s^4+32s}{4s^2+6})}{\lbrack(4s^2+6)^{-1}\rbrack^2} \\ =\frac{\frac{1}{4s^2+6}(\frac{12s^4+18s^2+8s^4+32s}{4s^2+6})}{\lbrack(4s^2+6)^{-1}\rbrack^2} \\ =\frac{\frac{1}{4s^2+6}(\frac{20s^4+18s^2+32s}{4s^2+6})}{\lbrack(4s^2+6)^{-1}\rbrack^2} \\ =\frac{\frac{20s^4+18s^2+32s}{(4s^2+6)^2}}{\frac{1}{(4s^2+6)^2}} \\ =\frac{(4s^2+6)^2(20s^4+18s^2+32s)}{(4s^2+6)^2} \\ =20s^4+18s^2+32s \end{gathered}[/tex]therefore the derivative is:
[tex]\frac{dg}{ds}=20s^4+18s^2+32s[/tex](Notice how we get the same result as before)
