given g(2)=3, g'(2)=-2, h(2)=-1, h'(2)=4 a) if f(x)=2g(x)+h(x), find f'(2) b) if f(x)=4-h(x), find f'(2) c) if f(x)=g(x)/h(x), find f'(2) d) if f(x)=g(x)h(x), find f'(2)

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given g(2)=3, g'(2)=-2, h(2)=-1, h'(2)=4 a) if f(x)=2g(x)+h(x), find f'(2) b) if f(x)=4-h(x), find f'(2) c) if f(x)=g(x)/h(x), find f'(2) d) if f(x)=g(x)h(x), find f'(2)

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for each, first find \(f'(x)\), then plugin \(x=2\)
I am still confused, can you walk me through a and I'll see if I can do the rest on my own?

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\[\large\rm f(x)=2g(x)+h(x)\]The derivative of f(x), with respect to x, is f'(x), yes?
Likewise, the derivative of g(x), with respect to x, is g'(x). There is nothing fancy going on in this first problem. No product rule, no composition, nothing special :)\[\large\rm f'(x)=2g'(x)+h'(x)\]
oh, and then you would plug in the given values of g'(x) and h'(x)?
\[\large\rm f'(\color{orangered}{x})=2g'(\color{orangered}{x})+h'(\color{orangered}{x})\]They want us to evaluate this at x=2,\[\large\rm f'(\color{orangered}{2})=2g'(\color{orangered}{2})+h'(\color{orangered}{2})\]And yes, use that chart at the top to plug in the missing pieces on the right :)
great! so it would be 2(-2)+4, so 0
Good!

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