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Mathematics 6 Online
OpenStudy (3psilon):

How to find the derivative of g(x) = 6x^2 / 2-x

OpenStudy (3psilon):

My teacher says you don't need the quotient rule to do it

OpenStudy (anonymous):

Using the product and chain rules, this is rather simple: We know: \[ g'(x)=\frac{d}{dx}(6x^2)(2-x)^{-1}=(12x)(2-x)^{-1}+6x^2\left(\frac{d}{dx}(2-x)^{-1}\right) \] From this, you should be able to get it.

OpenStudy (campbell_st):

ok... so your teacher uses the product rule... rewrite the problem as \[y = 6x^2 \times (2 -x)^{-1}\] and work from there

OpenStudy (3psilon):

Ahhhh I knew there was a way to write it in product form . Just forgot . Thanks

OpenStudy (3psilon):

@LolWolf would d/dx of \[(2-x)^{-1}\] just be -1? Sorry I'm new to derivatives

OpenStudy (anonymous):

Remember that we can us the chain rule to specify that \[\frac{d}{dx}f(g(x))=f'(g(x))g'(x)\]So, we apply this to show:\[\frac{d}{dx}(2-x)^{-1}=-1(2-x)^{-2}(-1)=(2-x)^{-2}\]And we are done.

OpenStudy (anonymous):

And, it's all right, if you need any further help, PM me.

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