How do you differentiate $z=A/y^10+Be^y$ ?

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Answer 1 · 2016-11-25T20:46:31

$dz/dy=-A/(10n^11)+Be^y$

Assuming $z$ is a function and $A$ and $B$ are constants, we see that:

$dz/dy=d/dy(A/y^10)+d/dy(Be^y)$

Pulling out the constants:

$dz/dy=Ad/dy(y^-10)+Bd/dy(e^y)$

Now using $d/dy(y^n)=ny^(n-1)$ and $d/dy(e^y)=e^y$ , we see that:

$dz/dy=A(-10n^(-10-1))+B(e^y)$

$dz/dy=-(10A)/(n^11)+Be^y$

How do you differentiate z=A/y^10+Be^yz=A/y^10+Be^y?