How do you find the derivative of $F (x) = \frac{1}{x - 5}$ $F(x) = 1/(x-5)$ ?

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Answer 1 · 2016-03-18T08:45:32

$\frac{d F}{d x} = \frac{- 1}{{(x - 5)}^{2}}$ $(dF)/(dx)=(-1)/(x-5)^2$

Let
$u = 1$ $u=1$
$v = x - 5$ $v=x-5$

We know, Quotient Rule

$\frac{v \frac{d}{d x} u - u \frac{d}{d x} v}{v^{2}}$ $[vd/dxu-ud/dxv]/v^2$

$\frac{(x - 5) \frac{d}{d x} 1 - 1 \frac{d}{d x} (x - 5)}{{(x - 5)}^{2}}$ $[(x-5)d/dx1-1d/dx(x-5)]/(x-5)^2$

$\frac{(x - 5) \cdot 0 - 1 \cdot 1}{{(x - 5)}^{2}} = \frac{- 1}{{(x - 5)}^{2}}$ $[(x-5)*0-1*1]/(x-5)^2=(-1)/(x-5)^2$

This will be the answer of the differentiated function.

How do you find the derivative of F(x)=1x−5F(x) = 1/(x-5)?