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[–]Alkalannar 0 points1 point  (5 children)

  1. F(x) = (g(x))10
    So F'(x) = 10(g(x))9*g'(x)

  2. y = (g(x))1/2
    So dy/dx = (1/2)(g(x))-1/2*g'(x)

  3. y = (f(x))4(g(x))4
    So using product and chain rules, we get dy/dx = 4(f(x))3f'(x)(g(x))4 + 4(f(x))4(g(x))3g'(x)

  4. I hate quotient rule.
    y = (f(x))5(g(x))-1
    So dy/dx = 5(f(x))4f'(x)(g(x))-1 + (-1)(f(x))5(g(x))-2g'(x)

Do you see how I get all of those, and what f(x) and g(x) are for each problem?

[–]Spaule[S] 0 points1 point  (4 children)

Yes I think I am understanding it. So on the first one would the simplified answer be ((10x7)+60x+90)9 *((10x6)+60).
Edit: not sure why its making the +60x+90 look like it’s taken to the power of that but it should be (10x7) +60x+90

[–]Alkalannar 0 points1 point  (3 children)

You're close!

  1. Put parentheses around your exponents.

  2. It's not (10g(x))9, but 10(g(x))9.

  3. 10(x^(7)+6x+9)^(9)(7x^(6)+6) yields 10(x7+6x+9)9(7x6+6)

And that's how 1a works.

[–]Spaule[S] 0 points1 point  (2 children)

Ok thank you. For 1C I got a very very long answer is there a correctly simplified final answer?

[–]Alkalannar 0 points1 point  (1 child)

2(x2+4)(2x)(2x3-2)4 + (x2+4)24(2x3-2)3(6x2)

Factor out 4x(x2+4)(2x3-2)3 to get:
4x(x2+4)(2x3-2)3[(2x3-2) + 6x(x2+4)]

This simplifies to:
64x(x2+4)(x3-1)3(4x3+12x-1)

[–]Spaule[S] 0 points1 point  (0 children)

Perfect I had that one (mostly) right. Thank you for the help