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Mathematics 9 Online
OpenStudy (anonymous):

derivative: x^(x+1)^1/2

OpenStudy (anonymous):

\[x ^{\sqrt{x+1}}\]

OpenStudy (apoorvk):

take log. then differentiate using chain rule and product rule.

OpenStudy (anonymous):

ok thanks

OpenStudy (anonymous):

\[\sqrt{x+1}/x + lnx/2\sqrt{x+1}\]

OpenStudy (anonymous):

is that right?

OpenStudy (anonymous):

derivative: x^(x+1)^1/2 (1/y)dy/dx=(x+1)^1/2 (x)^-1 +lnx/2(x+1)^1/2 dy/dy= y[(x+1)^1/2 (x)^-1 +lnx/2(x+1)^1/2], but y=x^(x+1)^1/2 dy/dx=[x^(x+1)^1/2] [(x+1)^1/2 (x)^-1 +lnx/2(x+1)^1/2] ans..

OpenStudy (anonymous):

did you get it azalea?

OpenStudy (anonymous):

your answer is correct except you need to multiply them by y=x^(x+1)^1/2

OpenStudy (anonymous):

do you want me to show you?

OpenStudy (anonymous):

thank you! can u please show me.

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