derivative of `y= (x^-1)(sqrtu)`

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derivative of `y= (x^-1)(sqrtu)`

Postby barnamah » Fri 08 April, 2011 15:00

how can I find the derivatives of`y= (x^-1)(sqrtu)` , u=3-5x
how can we do it without adding in the u and using the chain rule?
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Re: derivative of `y= (x^-1)(sqrtu)`

Postby barnamah » Fri 08 April, 2011 15:08

to take the derivatives of `y= (x^-1)(sqrtu)` using chain rule we can rewrite it as `y= (x^-1)(u^(1/2))`
we have two values multiplied so we can use chain rule.
`y'=-x^((-1-1))*u^(1/2)+(x^-1)*(1/2u^((1/2-1)))`
`y'=-x^(-2)*u^(1/2)+(x^-1)*(1/2u^((-1/2)))`
`y'=-x^(-2)*u^(1/2)+1/2x^-1*u^(-1/2)`

`y'=-x^-2sqrtu+1/2x^-1(1/sqrtu)`



that is the final answer.
but if we sub in the u=3-5x into equation we have:
`y'=-x^-2sqrt(3-5x)+1/2x^-1(1/sqrt(3-5x))`
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