How do you use the definition of a derivative to find the derivative of #f(x) = 4 / sqrt( 5 - x )#?

Question

Ezra Adams · Accepted Answer

#d/dx( 4/sqrt(5-x)) =2/((5-x)^(3/2)#

Using the definition of derivative:

#(df)/dx = lim_(h->0) (f(x+h)-f(x))/h#

we have:

#d/dx( 4/sqrt(5-x)) = lim_(h->0) 1/h(4/sqrt(5-x-h) - 4/sqrt(5-x))#

#d/dx( 4/sqrt(5-x)) = lim_(h->0) 4/h(sqrt(5-x) -sqrt(5-x-h)) /(sqrt(5-x-h) sqrt(5-x))#

Rationalize the numerator by multiplying and dividing by: #(sqrt(5-x) + sqrt(5-x-h))# and using the algebraic identity: #(a-b)(a+b) = a^2-b^2#

#d/dx( 4/sqrt(5-x)) = lim_(h->0) 4/h(sqrt(5-x) -sqrt(5-x-h)) /(sqrt(5-x-h) sqrt(5-x)) xx (sqrt(5-x) + sqrt(5-x-h))/(sqrt(5-x) + sqrt(5-x-h))#

#d/dx( 4/sqrt(5-x)) = lim_(h->0) 4/h((5-x) -(5-x-h)) /(sqrt(5-x-h) sqrt(5-x)(sqrt(5-x) + sqrt(5-x-h))#

#d/dx( 4/sqrt(5-x)) = lim_(h->0) 4/h(cancel(5)-cancel(x) -cancel(5)+cancel(x)+h) /(sqrt(5-x-h) sqrt(5-x)(sqrt(5-x) + sqrt(5-x-h))#

#d/dx( 4/sqrt(5-x)) = lim_(h->0) 4/cancel(h)cancel(h) /(sqrt(5-x-h) sqrt(5-x)(sqrt(5-x) + sqrt(5-x-h))#

#d/dx( 4/sqrt(5-x)) = 4/(sqrt(5-x) sqrt(5-x)(sqrt(5-x) + sqrt(5-x))#

#d/dx( 4/sqrt(5-x)) =4/(2(5-x)sqrt(5-x) )#

#d/dx( 4/sqrt(5-x)) =2/((5-x)^(3/2)#

Isaiah Allen · Answer

undefined To find the derivative of $ f(x) = \frac{4}{\sqrt{5 - x}} $, you can use the definition of a derivative. The definition states that the derivative of a function $ f(x) $ at a point $ x = a $ is the limit of the difference quotient as $ h $ approaches 0:

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

For $ f(x) = \frac{4}{\sqrt{5 - x}} $, you first need to rewrite the function as a power function:

$$ f(x) = 4(5 - x)^{-1/2} $$

Then, you can use the power rule for derivatives and the chain rule:

$$ f'(x) = \frac{d}{dx} [4(5 - x)^{-1/2}] $$
$$ f'(x) = 4 \cdot \frac{d}{dx} (5 - x)^{-1/2} $$
$$ f'(x) = 4 \cdot \frac{-1}{2}(5 - x)^{-3/2} \cdot (-1) $$

Simplify:

$$ f'(x) = 2(5 - x)^{-3/2} $$

So, the derivative of $ f(x) = \frac{4}{\sqrt{5 - x}} $ is $ f'(x) = 2(5 - x)^{-3/2} $.