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How to I do part (ii)?

How To I Do Part Ii class=

Sagot :

Answer:

Step-by-step explanation:

7(i)

[tex]\displaystyle\\\frac{d}{dx}\{x*(3x-5)^\frac{5}{3} \}=\frac{dx}{dx} *(3x-5)^\frac{5}{3}+x*\frac{d}{dx} \{(3x-5)^\frac{5}{3} \}\\\\ \frac{d}{dx}\{x*(3x-5)^\frac{5}{3} \}=1*(3x-5)^\frac{5}{3} +x*\frac{5}{3}*(3x-5)^{\frac{5}{3} -1}*\frac{d}{dx} \{(3x-5)\}\\\\ \frac{d}{dx}\{x*(3x-5)^\frac{5}{3} \}=(3x-5)^\frac{5}{3} +\frac{x*5*(3x-5)^{\frac{2}{3}}*3 }{3} \\\\\frac{d}{dx}\{x*(3x-5)^\frac{5}{3} \}=(3x-5)^\frac{5}{3} +5x*(3x-5)^\frac{2}{3} \\\\[/tex]

[tex]\displaystyle\\\frac{d}{dx}\{x*(3x-5)^\frac{2}{3} \}=(3x-5)^\frac{2}{3}*((3x-5)^{\frac{5}{3}-\frac{2}{3}} + 5x)\\\\\frac{d}{dx}\{x*(3x-5)^\frac{2}{3} \}=(3x-5)^\frac{2}{3} *(3x-5)^\frac{3}{3}+5x) \\\\\frac{d}{dx}\{x*(3x-5)^\frac{2}{3} \}=(3x-5)^\frac{2}{3} *(3x-5+5x)\\\ \frac{d}{dx}\{x*(3x-5)^\frac{2}{3} \}=(8x-5)*(3x-5)^\frac{2}{3}[/tex]

7(ii)

[tex]\int\limits {(x*(3x-5)^\frac{2}{3}) } \, dx \\Let\ \sqrt[3]{3x-5} =u\\Hence,\\(\sqrt[3]{3x-5})^3=u^3\\ 3x-5=u^3\\3x-5+5=u^3+5\\3x=u^3+5\\[/tex]

Divide both parts of the equation by 3:

[tex]\displaystyle\\x=\frac{u^3+5}{3}\\\\dx=\frac{d}{du} \{\frac{u^3+5}{3}\} \\dx=\frac{3u^2}{3} du=\\\\dx=u^2du\\Hence,\\\\\int\limits {\frac{u^3+5}{3}* u^2*u^2} \, du =\\\\\int\limits {\frac{u^3+5}{3} *u^4} \, du =\\\\\int\limits {\frac{u^7+5u^4}{3} } \, du =\\\\\frac{1}{3} \int\limits {u^7} \, du +\frac{1}{3}\int\limits {5u^4} \, dxu =\\\\[/tex]