How solve it? Derivate This Please!!!!

Question

1615 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-17T12:24:59

$(dF)/(dx)=(3(10x^3-7x^2)^2(60x^3-321x^2+140x))/(3x-10)^4$

As $F(x)=((10x^3-7x^2)/(3x-10))^3$ , let $G(x)=(10x^3-7x^2)/(3x-10)$

then we can use chain rule as $F(x)=[G(x)]^3$ .

But before that let us workout $(dG)/(dx)$

using Quotient rule, which states if $f(x)=(u(x))/(v(x))$

then $(df)/(dx)=((du)/(dx)xxv(x)-(dv)/(dx)xxu(x))/(v(x))^2$

In $G(x)=(10x^3-7x^2)/(3x-10)$ , we have $u(x)=10x^3-7x^2$ and $v(x)=3x-10$ .

and $(du)/(dx)=30x^2-14x$ and $(dv)/(dx)=3$

Hence, $(dG)/(dx)=((30x^2-14x)xx(3x-10)-3xx(10x^3-7x^2))/(3x-10)^2$

= $((90x^3-300x^2-42x^2+140x)-(30x^3-21x^2))/(3x-10)^2$

= $(60x^3-321x^2+140x)/(3x-10)^2$

Now $(dF)/(dx)=3(G(x))^2xx(dG)/(dx)$

= $3((10x^3-7x^2)/(3x-10))^2xx(60x^3-321x^2+140x)/(3x-10)^2$

= $(3(10x^3-7x^2)^2(60x^3-321x^2+140x))/(3x-10)^4$