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Hello, I've got this really tough multi-step problem that I just can't get. I've gotten the answer key, but just can't get the steps right. Can you help?

It is:

Suppose that f(x) and g(x) are differentiable everywhere and that f(4)=6, f'(4)=3, g(4)=-2 and g'(4)=10

Part 1: if y=2f(x)g(x), what is y'(4)

Part 2: if y=g(x)/f(x), what is y'(4)

Part 3: Find the equation of the tangent line to 3(f(x))^2 at x=4

Thank you for your assistance,

Jenny

Part 1: If y(x) = 2f(x)g(x), then y'(x) = 2[f'(x)g(x) + g(x)'f(x)].

This is an example of the product rule. That states when y = fg, then y' = fg' + gf'.

At x=4, this is y'(4) = 2[f'(4)g(4) + g'(4)f(4)].

By what we're told, that is that is y'(4) = 2[3(-2) + 10*6].

That works out to be y'(4) = 2(-6 + 60) = 2(54) = 108.

Part 2: If y = f(x)/g(x), that uses the quotient rule.

That is, y' = (g(x)f'(x) - f(x)g'(x))/g²(x).

At x=4, that is given as (-2*3 - 6*10)/(-2)².

That works out to (-6 - 60)/4 = -66/4 = -16.5.

Part 3: Find the tangent to 3 * f²(x).

If y = 3 * f²(x), the slop is the derivative. The derivative of this function is

y' = 6*f(x)f'(x).

That is given in the data as y' = 6*6*3 = 108.

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