Home > Calculus > Instantaneous Rate of Change at a Point

How to find instantaneous rate of change for #y(x)=1/(x+2)# at x=2?

Answer 1

Samantha Adkison

#-1/16#

At x = 2, y(x)'s instantaneous rate of change is y'(2).

#"Express "y(x)=1/(x+2)=(x+2)^-1#

differentiate using the #color(blue)"chain rule"#

#rArry'(x)=-(x+2)^-2 .d/dx(x+2)#

#rArry'(x)=-(x+2)^-2=-1/(x+2)^2#

#rArry'(2)=-1/(4)^2=-1/16#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Axel Alvarez

To find the instantaneous rate of change for ( y(x) = \frac{1}{x+2} ) at ( x = 2 ), we can use the concept of the derivative.

The derivative of a function represents the rate at which the function is changing at any given point.

To find the derivative of ( y(x) ), we'll use the power rule for differentiation:

[ \frac{d}{dx} \left( \frac{1}{x+2} \right) = -\frac{1}{{(x+2)}^2} ]

Now, we can evaluate the derivative at ( x = 2 ) to find the instantaneous rate of change:

[ y'(x) = -\frac{1}{{(2+2)}^2} = -\frac{1}{16} ]

So, the instantaneous rate of change for ( y(x) = \frac{1}{x+2} ) at ( x = 2 ) is ( -\frac{1}{16} ).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.