Question

Find the differential dy of y=1/x+1 Evaluate dy for x=1 , dx=−.01

Answer 1

Assuming y=1/(x+1)
y'=-1/(x+1)^2
dy=-1/(x+1)^2*dx
=-1/(1+1)^2*(-.01)
=0.0025

Answer 2

Value of differential function $dy = 0.0025$ for the given $y = \frac{1}{x+1}$.

What is differentiation?

" Differentiation is defined as the rate of change of dependent variable with respect to the independent variable."

Formula used

$\frac{d}{dx} (\frac{1}{x^{n} } )= \frac{-n}{x^{n+1} }$

According to the question,

Given function,

$y = \frac{1}{x+1}$

Differentiate it with respect to $'x'$  using the formula we get,

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

Evaluate the differential function $dy$ for  $x=1 , dx =- .01$ we get,

$dy = \frac{-1}{(x+1)^{2} } dx\\\\\implies dy = \frac{-1}{(1+1)^{2} } (-0.01)\\\\\implies dy = \frac{0.01}{(4) }\\\\\implies dy = 0.0025$

Hence, the value of the differential function $dy = 0.0025$ for the given $y = \frac{1}{x+1}$.

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