What is the derivative of $\frac{x + 1}{y}$ $(x + 1)/y$ ?

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Answer 1 · 2015-08-15T18:12:09

$\frac{d}{d x} (\frac{x + 1}{y}) = \frac{y - (x + 1) \frac{d y}{d x}}{y^{2}}$ $d/dx((x+1)/y) = ( y-(x+1) dy/dx)/y^2$

$\frac{d}{d x} (\frac{x + 1}{y}) = \frac{(1) y - (x + 1) \frac{d y}{d x}}{y^{2}}$ $d/dx((x+1)/y) = ((1)y-(x+1) dy/dx)/y^2$

There's not much more we can do without the rest of the equation relating $x$ $x$ and $y$ $y$ .

We could rewrite a bit, to get:

$= \frac{y - (x + 1) \frac{d y}{d x}}{y^{2}}$ $= ( y-(x+1) dy/dx)/y^2$

or

$= \frac{1}{y} - \frac{x + 1}{y^{2}} \frac{d y}{d x}$ $= 1/y - (x+1)/y^2 dy/dx$

What is the derivative of (x + 1)/yx+1y(x + 1)/y?