How do you use the limit definition of the derivative to find the derivative of #f(x)=sqrt(-3x-5)#?

1 Answer
Steve M
Nov 6, 2017

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

Explanation:

We seek:

# d/dx sqrt(-3x-5) #

By First Principles, using the limit definition:

# f'(x) = lim_(h rarr 0) (f(x+h) - f(x))/h #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) (sqrt(-3(x+h)-5) - sqrt(-3x-5))/h #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) (sqrt(-3x-3h-5) - sqrt(-3x-5))/h *(sqrt(-3x-3h-5) + sqrt(-3x-5))/(sqrt(-3x-3h-5) + sqrt(-3x-5))#

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) ((sqrt(-3x-3h-5) - sqrt(-3x-5)) (sqrt(-3x-3h-5) + sqrt(-3x-5))) / (h(sqrt(-3x-3h-5) + sqrt(-3x-5)))#

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( sqrt(-3x-3h-5)^2 - sqrt(-3x-5)^2 ) / (h(sqrt(-3x-3h-5) + sqrt(-3x-5)))#

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( (-3x-3h-5) - (-3x-5) ) / (h(sqrt(-3x-3h-5) + sqrt(-3x-5)))#

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( -3x-3h-5 +3x+5 ) / (h(sqrt(-3x-3h-5) + sqrt(-3x-5)))#

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( -3h ) / (h(sqrt(-3x-3h-5) + sqrt(-3x-5)))#

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) ( -3 ) / (sqrt(-3x-3h-5) + sqrt(-3x-5))#

# \ \ \ \ \ \ \ \ = ( -3 ) / (sqrt(-3x+0-5) + sqrt(-3x-5))#

# \ \ \ \ \ \ \ \ = ( -3 ) / (sqrt(-3x-5) + sqrt(-3x-5))#

# \ \ \ \ \ \ \ \ = ( -3 ) / (2sqrt(-3x-5))#

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