PLEASE HELP NOW BASED OFF THE BOOK "The Giver" THXS IN ADVANCE

PLEASE HELP NOW BASED OFF THE BOOK The Giver THXS IN ADVANCE

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carapiasebas

04.04.2022

Mathematics

Remember
chain rule: derivitive of f(g(x))=f'(g(x))g'(x)
power rule: f(x)^n=nf(x)^(n-1)f'(x)
quotient rule: derivitive of g(x)/f(x)=(g'(x)f(x)-f'(x)g(x))/((g(x))^2)

so

first chain rule/power rule
$5( \frac{5x^5-3}{2x^3-3})^4 (\frac{5x^5-3}{2x^3-3}) \frac{dy}{dx}$ =
first, (5x^5-3)dy/dx=25x^4 and (2x^3-3)dy/dx=6x^2
$5( \frac{5x^5-3}{2x^3-3})^4 (\frac{(5x^5-3)dy/dx(2x^3-3)-(2x^3-3)dy/dx(5x^5-3)}{(2x^3-3)^2})$ =
$5( \frac{5x^5-3}{2x^3-3})^4 (\frac{(25x^4)(2x^3-3)-(6x^2)(5x^5-3)}{(2x^3-3)^2})$ =
$5( \frac{5x^5-3}{2x^3-3})^4 (\frac{50x^7-75x^4-(30x^7-18x^2)}{(2x^3-3)^2})$ =
$5( \frac{5x^5-3}{2x^3-3})^4 (\frac{50x^7-75x^4-30x^7+18x^2)}{(2x^3-3)^2})$ =
$5( \frac{5x^5-3}{2x^3-3})^4 (\frac{20x^7-75x^4+18x^2)}{(2x^3-3)^2})$
simplify further if you want

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