The derivative of Algebraic Functions

The derivative of Algebraic Functions

If you could still remember your first derivative from the previous topic on functions, this is the follow-up. It's a principle almost the same except that this page deals with a bit higher.

The formula goes the same, see fig. I below.

Fig. I

Illustrative Examples:

Perform the indicated operation. (Problems can be found from the book Differential and Integral Calculus by Love and Rainville).

$1.) w = \frac{1}{6} +\frac{2}{5}{\color{Red} y} -\frac{{\color{Red} y}^{2}}{3}$

${w}' = 0 +\frac{2}{5}{\color{Red} y^{1-1}} -\frac{{\color{Red} y}^{{\color{Red} 2-1}}}{3}$

${w}' = \frac{2}{5}{\color{Red} y}^{{\color{Red} 0}} -\frac{{\color{Red} y}}{3}$

${w}' = \frac{2}{5} -\frac{{\color{Red} y}}{3}$

$2.)w=\frac{2}{3} -\frac{{\color{Red} y}^{2}}{5} +\frac{4}{3}{\color{Red} y}^{3}$

${w}'= 0 -\frac{2{\color{Red} y^{2-1}}}{5} +{\color{Blue} 3}(\frac{4}{{\color{Blue} 3}}{\color{Red} y^{3-1}})$

${w}'= -\frac{2{\color{Red} y^{1}}}{5} +{\color{Blue} 3}(\frac{4}{{\color{Blue} 3}}{\color{Red} y^{2}})$

${w}'= -\frac{2}{5} {\color{Red} y}+{}({4}{{}}{\color{Red} y^{2}})$

$3.)y=\sqrt{{\color{Red} x}}-\frac{4}{\sqrt{{\color{Red} x}}}$

$y={\color{Red} x}^{\frac{1}{2}} -4{\color{Red} x}^{\frac{1}{2}}$

${y}'=\frac{1}{2}{\color{Red} x}^{\frac{1}{2}-1} - \frac{1}{2}{\color{Red} x}^{-\frac{1}{2}-1}$

${y}'=\frac{1}{2}{\color{Red} x}^{\frac{1}{2}-1} -2{\color{Red} x}^{-\frac{3}{2}}$

${y}'=\frac{1}{2{\sqrt{{\color{Red} x}}}}- \frac{2}{{\sqrt{{\color{Red} x}^{3}}}}$

Sample Problems:

Perform the indicated operation. (Problems can be found from the book Differential and Integral Calculus by Love and Rainville).

$1.) y=\frac{2}{^{\sqrt{{\color{Red} x}}}} +\frac{4}{^{\sqrt{{\color{Red} x}^{3}}}}=2{\color{Red} x}^{-\frac{1}{2}} +4{\color{Red} x}^{-\frac{3}{2}}$

${y}'=-\frac{1}{{\color{Blue} 2}}({\color{Blue} 2}){\color{Red} x}^{-\frac{1}{2}-1} -\frac{3}{2}(4){\color{Red} x}^{-\frac{3}{2}-1}$

${y}'=-{\color{Red} x}^{-\frac{3}{2}} -6{\color{Red} x}^{-\frac{5}{2}}$

${y}'= -\frac{1}{^{\sqrt{{\color{Red} x}^{3}}}} -\frac{6}{^{^{\sqrt{{\color{Red} x}^{5}}}}}$

${y}'= -\frac{1}{{\color{Red} x}{\sqrt{{\color{Red} x}}}} -\frac{6}{{\color{Red} x}^{2}\sqrt{{\color{Red} x}}}$

$2.) x=\sqrt{{\color{Red} t}} +\frac{-1}{\sqrt{{\color{Red} t}}}$

$x={\color{Red} t}^{\frac{1}{2}} -{\color{Red} t}^{-\frac{1}{2}}$

${x}'= \frac{1}{2}{\color{Red} t}^{\frac{1}{2}-1} +\frac{1}{2}{\color{Red} t}^{-\frac{1}{2}-1}$

${x}'= \frac{1}{2}{\color{Red} t}^{-\frac{1}{2}} +\frac{1}{2}{\color{Red} t}^{-\frac{3}{2}}$

${x}'= \frac{1}{2{\sqrt{{\color{Red} t}}}} +\frac{1}{2{\sqrt{{\color{Red} t}^{3}}}}$

$3.) x=2{\color{Red} t}\sqrt{{\color{Red} t}}+\frac{1}{\sqrt{{\color{Red} t}}}$

$x=2{\color{Red} t}({\color{Red} t}^{\frac{1}{2}}) +{\color{Red} t}^{-\frac{1}{2}}$

${x}'=2{\color{Red} t}^{}\frac{3}{2} +{\color{Red} t}^{-\frac{1}{2}}$

${x}'=\frac{3}{{\color{Blue} 2}}({\color{Blue} 2}){\color{Red} t}^{\frac{3}{2}-1} -\frac{1}{2}{\color{Red} t}^{-\frac{1}{2}-1}$

${x}'=3{\color{Red} t}^{\frac{1}{2}} -\frac{{\color{Red} t}}{2}^{-\frac{3}{2}}$

${x}'=3{\sqrt{{\color{Red} t}}} -\frac{1}{2{\sqrt{{\color{Red} t}^{3}}}}$

${x}'=3{\sqrt{{\color{Red} t}}} -\frac{2}{2{\sqrt{{\color{Red} t}^{2}}{\sqrt{{\color{Red} t}}}}}$

${x}'= 3{\sqrt{{\color{Red} t}}} -\frac{2}{2{\color{Red} t}{\sqrt{{\color{Red} t}}}}$

$4.) y=6 -\frac{1}{\sqrt[3]{{\color{Red} x}^{4}}}$

$y=6 - {\color{Red} x}^{-\frac{4}{3}}$

${ y}'=0 +\frac{4}{3}({\color{Red} x}^{-\frac{4}{3}})$

${ y}'=\frac{4}{3}{\color{Red} x}^{-\frac{7}{3}}$

${ y}'=\frac{4}{\sqrt[3]{{\color{Red} x}^{7}}}$

${ y}'=\frac{4}{{\sqrt[3]{{\color{Red} x}^{6}}}{\sqrt[3]{{\color{Red} x}}}}$

${ y}'=\frac{4}{{\color{Red} x}^{2}\sqrt{{\color{Red} x}}}$

$5.)y=6\sqrt[3]{{\color{Red} x}} -{\color{Red} x}^{3}= 6({\color{Red} x}^{\frac{1}{3}}) -{\color{Red} x}^{3}$

${y}'= \frac{1}{3}(6{\color{Red} x}^{\frac{1}{3}-1}) -3{\color{Red} x}^{3-1}$

${y}'= 2{\color{Red} x}^{-\frac{2}{3}} -3{\color{Red} x}^{2}$

${y}'= \frac{2}{^{\sqrt[3]{{\color{Red} x}}}} -3{\color{Red} x}^{2}$

Seatwork with a partial solution.

Perform the indicated operation.

$1.)y=\frac{4}{\sqrt{{\color{Red} x}^{3}}}+\frac{2}{\sqrt{{\color{Red} x}}}$

$y=4({\color{Red} x}^{3})^{-\frac{1}{2}}+2({\color{Red} x}^{-\frac{1}{2}})$

$y=4({\color{Red} x})^{-\frac{3}{2}}+2({\color{Red} x}^{-\frac{1}{2}})$

Click here to finalize your answer

$2.)y= 3\sqrt{{\color{Red} x}} +\frac{1}{\sqrt{{\color{Red} x}}}$

$y=3{\color{Red} x}^{\frac{1}{2}}+{\color{Red} x}^{-\frac{1}{2}}$

$y{}'=$ Click here to finalize your answer

$3.) y=\frac{3{\color{Red} x}}{\sqrt{{\color{Red} x}}}+\sqrt{{\color{Red} x}}$

$y=3{\color{Red} x}({\color{Red} x}^{-\frac{1}{2}})+{\color{Red} x}^\frac{1}{2}$

$y{}'=$ Click here to finalize your answer

QUIZ

Find the first derivative by performing the indicated operation.

$1.) y= \frac{3-{\color{Red} t}^{2}}{\sqrt{{\color{Red} t}}}$

$2.) x= \frac{2\sqrt{{\color{Red} t}-1}}{^{\sqrt[3]{{\color{Red} t}}}}$

$3.) y= {\color{Red} x}{\sqrt[3]{{\color{Red} x}}}-\frac{2{\color{Red} x}}{\sqrt[3]{{\color{Red} x}}}$

$4.) y= \frac{({\color{Red} v}^{2}-4)}{({\color{Red} v}^{2}-3)}$

$5.) y=\frac{2}{3{\color{Red} x}^{2}-1}$

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DMG

The derivative of Algebraic Functions

Posted by : Allan_Dell on Friday, April 10, 2020 | 3:45 PM

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DMG

The derivative of Algebraic Functions

Posted by : Allan_Dell on Friday, April 10, 2020 | 3:45 PM

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Posted by : Allan_Dell on Friday, April 10, 2020 | 3:45 PM