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OpenStudy (anonymous):

If y^5+xy=2x-4, find dy/dx

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OpenStudy (anonymous):

ok why not? the derivative of \[f^5(x)\] is \[5f^4(x)\times f'(x)\] by the chain rule, meaning the derivative of \[y^5\] is \[5y^4y'\] as above

OpenStudy (anonymous):

the derivative of \[xf(x)\] is \[f(x)+xf'(x)\] by the product rule, so the derivative of \[xy\] is \[y+xy'\] and so on

OpenStudy (anonymous):

Confused again...

OpenStudy (anonymous):

oh! Okay! I see...so how do i do the other half of the equation?

OpenStudy (anonymous):

for the xy part, use product rule

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OpenStudy (anonymous):

1.y+xy'

OpenStudy (anonymous):

for the right hand side =2

OpenStudy (anonymous):

if u combine all these u get 5y^4y'+y+xy'=2 y'(5y^4+x)=2-y y'=2-y/5y^4+x

OpenStudy (anonymous):

Thank you! That was the perfect reply...

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