How do you find the derivative of #f(x)=4/(sqrtx-5)# using the limit definition?

Question

Christopher Allers · Accepted Answer

#f'(x)=(-2)/((sqrt(x)-5)(sqrt(x)-5)(sqrt(x)))# #f'(x)=lim_(hrarr0)(4/(sqrt(x+h)-5)-4/(sqrt(x)-5))/h# (difference quotient) #f'(x)=lim_(hrarr0)((4(sqrt(x)-5))/((sqrt(x+h)-5)(sqrt(x)-5))-(4(sqrt(x+h)-5))/((sqrt(x+h)-5)(sqrt(x)-5)))/h# (combine numerator into 1 fraction) #f'(x)=lim_(hrarr0)(4sqrt(x)-20-4sqrt(x+h)+20)/((sqrt(x+h)-5)(sqrt(x)-5)h)# (multiply denominator of the fraction in the numerator to the denominator of the larger fraction) #f'(x)=lim_(hrarr0)(4sqrt(x)-4sqrt(x+h))/((sqrt(x+h)-5)(sqrt(x)-5)h)# #f'(x)=lim_(hrarr0)((4sqrt(x)-4sqrt(x+h))(4sqrt(x)+4sqrt(x+h)))/((sqrt(x+h)-5)(sqrt(x)-5)h(4sqrt(x)+4sqrt(x+h)))# (conjugate) #f'(x)=lim_(hrarr0)(16x-16x-16h)/((sqrt(x+h)-5)(sqrt(x)-5)h(4sqrt(x)+4sqrt(x+h)))# #f'(x)=lim_(hrarr0)(-16h)/((sqrt(x+h)-5)(sqrt(x)-5)h(4sqrt(x)+4sqrt(x+h)))# #f'(x)=lim_(hrarr0)(-16)/((sqrt(x+h)-5)(sqrt(x)-5)(4sqrt(x)+4sqrt(x+h)))# (remove h from both numerator and denominator) now you can use direct substitution: #f'(x)=(-16)/((sqrt(x+0)-5)(sqrt(x)-5)(4sqrt(x)+4sqrt(x+0)))# #f'(x)=(-16)/((sqrt(x)-5)(sqrt(x)-5)(4sqrt(x)+4sqrt(x)))# #f'(x)=(-16)/((sqrt(x)-5)(sqrt(x)-5)(8sqrt(x)))# #f'(x)=(-2)/((sqrt(x)-5)(sqrt(x)-5)(sqrt(x)))#

Isaiah Allen · Answer

undefined To find the derivative of $ f(x) = \frac{4}{\sqrt{x} - 5} $ using the limit definition:

1. Begin with the function $ f(x) = \frac{4}{\sqrt{x} - 5} $.
2. Let $ h $ be a small change in $ x $, so $ x + h $ represents a point close to $ x $.
3. Find $ f(x + h) $ by substituting $ x + h $ into the function.
4. Calculate $ \frac{f(x + h) - f(x)}{h} $.
5. Take the limit as $ h $ approaches 0 of $ \frac{f(x + h) - f(x)}{h} $ to find the derivative.

Following these steps, you'll arrive at the derivative of the function $ f(x) = \frac{4}{\sqrt{x} - 5} $.