How do you use the limit definition to find the derivative of #f(x)=2/(x+4)#?

Question

Emma Aldridge · Accepted Answer

#f'(x)=(-2)/((x+4)^2)#

def of derivative

#f'(x)=limh->0(f(x+h)-f(x))/(h)#

Substitution

#f'(x)=limh->0(2/(x+h+4)-2/(x+4))/(h)#

Common Denominator

#f'(x)=limh->0((2(x+4))/((x+4)(x+h+4))-(2(x+h+4))/((x+4)(x+h+4)))/(h)#

Distribute and write as a single numerator

#f'(x)=limh->0((2x+8)/((x+4)(x+h+4))-(2x+2h+8)/((x+4)(x+h+4)))/(h)#

#f'(x)=limh->0((2x+8-2x-2h-8)/((x+4)(x+h+4)))/(h)#

Simplify

#f'(x)=limh->0((cancel(2x)cancel(+8)cancel(-2x)-2hcancel(-8))/((x+4)(x+h+4)))/(h)#

#f'(x)=limh->0((-2h)/((x+4)(x+h+4)))/(h)#

Multiply by the reciprocal

#f'(x)=limh->0(-2h)/((x+4)(x+h+4))*(1/h)#

#f'(x)=limh->0(-2h)/(h(x+4)(x+h+4))#

Simplify

#f'(x)=limh->0(-2cancelh)/(cancelh(x+4)(x+h+4))#

#f'(x)=limh->0(-2)/((x+4)(x+h+4))#

Now we can substitute in a 0 for h

#f'(x)=(-2)/((x+4)(x+0+4))#

Simplify

#f'(x)=(-2)/((x+4)(x+4))#

Simplify

#f'(x)=(-2)/((x+4)^2)#

Watch this tutorial to see a similar question solved used the same methods.

William Abdalla · Answer

undefined To use the limit definition to find the derivative of $ f(x) = \frac{2}{x+4} $, you first apply the definition of the derivative:

$$ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} $$

Substitute $ f(x) = \frac{2}{x+4} $ into the equation:

$$ f'(x) = \lim_{h \to 0} \frac{\frac{2}{x+h+4} - \frac{2}{x+4}}{h} $$

Simplify the expression:

$$ f'(x) = \lim_{h \to 0} \frac{2(x+4) - 2(x+h+4)}{h(x+4)(x+h+4)} $$

$$ f'(x) = \lim_{h \to 0} \frac{2x + 8 - 2x - 2h - 8}{h(x+4)(x+h+4)} $$

$$ f'(x) = \lim_{h \to 0} \frac{-2h}{h(x+4)(x+h+4)} $$

$$ f'(x) = \lim_{h \to 0} \frac{-2}{(x+4)(x+h+4)} $$

Now, evaluate the limit as $ h $ approaches 0:

$$ f'(x) = \frac{-2}{(x+4)^2} $$

So, the derivative of $ f(x) = \frac{2}{x+4} $ is $ f'(x) = \frac{-2}{(x+4)^2} $.