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OpenStudy (anonymous):

Find the Derivative, using Chain Rule. y=6x^2+5cosx

OpenStudy (kainui):

Since you say use the chain rule, does that mean you really have: \[\LARGE y=6x^{2+5\cos x}\]

OpenStudy (jdoe0001):

eheh... I don't see much chain ruling there, unless is an exponentional or otherwise

OpenStudy (kainui):

Which is why I'm asking...

OpenStudy (anonymous):

My Mistake, \[y=6x^2+5cosx\]

OpenStudy (anonymous):

You are right there is no chain rule

OpenStudy (jdoe0001):

the only thing I can see chain-ruling would be the \(x^2\) and the cos(x) other than that.... is just a constant in front

OpenStudy (jdoe0001):

so.. chain-rule those 2 I gather....

OpenStudy (anonymous):

6x^2 is done by the power rule.

OpenStudy (anonymous):

So, it would be 12x, but what about the 5cosx?

OpenStudy (anonymous):

What is the derivative of cos(x)?

OpenStudy (anonymous):

-sin?

OpenStudy (anonymous):

so it would be -5sinx?

OpenStudy (anonymous):

Yes

OpenStudy (jdoe0001):

well... one could say \(\bf 6\cdot \cfrac{d}{dx}[x^2]\cdot \cfrac{d}{dx}[x]\implies 6\cdot 2x^1\cdot 1x^0\implies 12x\)

OpenStudy (anonymous):

so my answer is 12x+-sin5x?

OpenStudy (jdoe0001):

well... you don't really need the chain-rule per se but I gather the exercise is to use it

OpenStudy (jdoe0001):

but yes... is 12x-5sin(x)

OpenStudy (jdoe0001):

to say chain-rule the cos(x) part... one could \(\bf 5cos(x)\implies 5\cdot \cfrac{d}{dx}[cos(x)]\cdot \cfrac{d}{dx}[x]\)

OpenStudy (anonymous):

Okay Thank You (:

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