Differentiate Calculus Help?

Question

I have some homework problems that I'm unable to work out.

1) Differentiate: (x^2 + 3) (x^2 -3)^10

2) Differentiate: x+2/(x+3) (x+4)

3) Differentiate: y=3^x +x^3

4) Differentiate: f(x)= ln( (3^x+7x-5)/8x+1)^4/5

Answer 1

1.) d/dx (x^2 + 3) (x^2 - 3)^10

Differentiation Product Rule: (fg)' = gf' + fg'
Define g and f, then define g' and f':
g = x^2 + 3
g' = 2x
f = (x^2 - 3)^10

Use the chain rule to find f':
Chain Rule: If f(x) = h(g(x)), f'(x) = h'(g(x)) * g'(x)
g(x) = x^2 - 3
g'(x) = 2x
h(x) = x^10
h'(x) = 10x^9
f'(x) = 20(x^2 - 3)^9 * 2x = 40x(x^2 - 3)^9

So, now that we have g, f, g', and f', we can put our answer together:
(fg)' = gf' + fg'
(fg)' = (x^2 + 3)(40x(x^2 - 3)^9) + ((x^2 - 3)^10)(2x)
(fg)' = 40x(x^2 + 3)(x^2 - 3)^9 + 2x(x^2 - 3)^10 (solution!)

2.) Use the Quotient Rule:
If f(x) = g(x) / h(x), then f'(x) = (g'(x)h(x) - g(x)h'(x))/h(x)^2

Define g(x), h(x), g'(x), h'(x), and h(x)^2:
g(x) = x + 2
h(x) = (x + 3)(x + 4) = x^2 + 7x + 12
g'(x) = 1
h'(x) = 2x + 7
h(x)^2 = (x+3)^2(x+4)^2

Plug it in and simplify:
f'(x) = (g'(x)h(x) - g(x)h'(x))/h(x)^2
f'(x) = ((1)(x+3)(x+4) - (x + 2)(2x + 7))/(x+3)^2(x+4)^2
f'(x) = ((x+3)(x+4) - (x + 2)(2x + 7))/(x+3)^2(x+4)^2

Split the fractions for the final answer:
f'(x) = ((x+3)(x+4)) / (x+3)^2(x+4)^2 - (x + 2)(2x + 7)/(x+3)^2(x+4)^2
f'(x) = 1 / (x+3)(x+4) - (x + 2)(2x + 7)/(x+3)^2(x+4)^2 (solution!)

3.) y=3^x +x^3
Use natural log to clear the exponent, then differentiate both sides:
ln y = x ln 3 + 3 ln x
d/dy ln y = 1/y
d/dx x ln 3 + 3 ln x = ln 3 + 3 / x

That gives you:
1/y dy = ln 3 + 3 / x dx
dy / dx = (ln 3 + 3 / x) / (1/y)
dy / dx = y ln 3 + 3y / x (solution)

4.) This one essentially combines all 3 methods above - but now that you can see how to work each method, you should be able to solve it.

Answer 2

well from what i know of differentiation the third one looks like it may be
f'(x) = 3x^x-1+3x^2