Question

How to I do part (ii)?

Answer 1

Explanation:

7(i)

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

`\displaystyle\\(d)/(dx)\{x*(3x-5)^(2)/(3) \}=(3x-5)^(2)/(3)*((3x-5)^{(5)/(3)-(2)/(3)} + 5x)\\\\(d)/(dx)\{x*(3x-5)^(2)/(3) \}=(3x-5)^(2)/(3) *(3x-5)^(3)/(3)+5x) \\\\(d)/(dx)\{x*(3x-5)^(2)/(3) \}=(3x-5)^(2)/(3) *(3x-5+5x)\\\ (d)/(dx)\{x*(3x-5)^(2)/(3) \}=(8x-5)*(3x-5)^(2)/(3)`

7(ii)

`\int\limits {(x*(3x-5)^(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\\`

Divide both parts of the equation by 3:

`\displaystyle\\x=(u^3+5)/(3)\\\\dx=(d)/(du) \{(u^3+5)/(3)\} \\dx=(3u^2)/(3) du=\\\\dx=u^2du\\Hence,\\\\\int\limits {(u^3+5)/(3)* u^2*u^2} \, du =\\\\\int\limits {(u^3+5)/(3) *u^4} \, du =\\\\\int\limits {(u^7+5u^4)/(3) } \, du =\\\\(1)/(3) \int\limits {u^7} \, du +(1)/(3)\int\limits {5u^4} \, dxu =\\\\`