How do you differentiate $z = \frac{A}{y^{10}} + B e^{y}$ $z=A/y^10+Be^y$ ?

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How do you differentiate $z = \frac{A}{y^{10}} + B e^{y}$ $z=A/y^10+Be^y$ ?

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Answer 1 · 2016-11-25T20:46:31

$\frac{d z}{d y} = - \frac{A}{10 n^{11}} + B e^{y}$ $dz/dy=-A/(10n^11)+Be^y$

Explanation:

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

$\frac{d z}{d y} = \frac{d}{d y} (\frac{A}{y^{10}}) + \frac{d}{d y} (B e^{y})$ $dz/dy=d/dy(A/y^10)+d/dy(Be^y)$

Pulling out the constants:

$\frac{d z}{d y} = A \frac{d}{d y} (y^{- 10}) + B \frac{d}{d y} (e^{y})$ $dz/dy=Ad/dy(y^-10)+Bd/dy(e^y)$

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

$\frac{d z}{d y} = A (- 10 n^{- 10 - 1}) + B (e^{y})$ $dz/dy=A(-10n^(-10-1))+B(e^y)$

$\frac{d z}{d y} = - \frac{10 A}{n^{11}} + B e^{y}$ $dz/dy=-(10A)/(n^11)+Be^y$

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How do you differentiate $z = \frac{A}{y^{10}} + B e^{y}$ $z=A/y^10+Be^y$ ?

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Explanation:

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