Answer :
Sure, I'll guide you through finding the derivative of [tex]\( f(x) = \sqrt{5x + 9} \)[/tex] using the limit definition of a derivative step by step.
### Question 1
Given [tex]\( f(x) = \sqrt{5x + 9} \)[/tex], let's find the derivative [tex]\( f'(x) \)[/tex] using the definition:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \][/tex]
Step 1: Write the function [tex]\( f(x + h) \)[/tex].
[tex]\[ f(x + h) = \sqrt{5(x + h) + 9} = \sqrt{5x + 5h + 9} \][/tex]
Step 2: Subtract [tex]\( f(x) \)[/tex] from [tex]\( f(x + h) \)[/tex].
[tex]\[ f(x + h) - f(x) = \sqrt{5x + 5h + 9} - \sqrt{5x + 9} \][/tex]
Step 3: Substitute this difference into the limit definition of the derivative:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{5x + 5h + 9} - \sqrt{5x + 9}}{h} \][/tex]
Step 4: Rationalize the numerator by multiplying and dividing by the conjugate of the numerator:
[tex]\[ \frac{\sqrt{5x + 5h + 9} - \sqrt{5x + 9}}{h} \cdot \frac{\sqrt{5x + 5h + 9} + \sqrt{5x + 9}}{\sqrt{5x + 5h + 9} + \sqrt{5x + 9}} = \lim_{h \to 0} \frac{(5x + 5h + 9) - (5x + 9)}{h (\sqrt{5x + 5h + 9} + \sqrt{5x + 9})}\][/tex]
Step 5: Simplify the numerator:
[tex]\[ = \lim_{h \to 0} \frac{5h}{h (\sqrt{5x + 5h + 9} + \sqrt{5x + 9})}\][/tex]
Step 6: Cancel out [tex]\( h \)[/tex] in the numerator and denominator:
[tex]\[ = \lim_{h \to 0} \frac{5}{\sqrt{5x + 5h + 9} + \sqrt{5x + 9}}\][/tex]
Step 7: As [tex]\( h \)[/tex] approaches 0, [tex]\( \sqrt{5x + 5h + 9} \)[/tex] approaches [tex]\( \sqrt{5x + 9} \)[/tex]:
[tex]\[ = \frac{5}{\sqrt{5x + 9} + \sqrt{5x + 9}} = \frac{5}{2\sqrt{5x + 9}}\][/tex]
### Final Answer
So, the derivative of the function [tex]\( f(x) = \sqrt{5x + 9} \)[/tex], expressed as the limit, is:
[tex]\[ f'(x) = \frac{5}{2\sqrt{5x + 9}} \][/tex]
This completes the step-by-step derivation using the definition of the derivative.
### Question 1
Given [tex]\( f(x) = \sqrt{5x + 9} \)[/tex], let's find the derivative [tex]\( f'(x) \)[/tex] using the definition:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \][/tex]
Step 1: Write the function [tex]\( f(x + h) \)[/tex].
[tex]\[ f(x + h) = \sqrt{5(x + h) + 9} = \sqrt{5x + 5h + 9} \][/tex]
Step 2: Subtract [tex]\( f(x) \)[/tex] from [tex]\( f(x + h) \)[/tex].
[tex]\[ f(x + h) - f(x) = \sqrt{5x + 5h + 9} - \sqrt{5x + 9} \][/tex]
Step 3: Substitute this difference into the limit definition of the derivative:
[tex]\[ f'(x) = \lim_{h \to 0} \frac{\sqrt{5x + 5h + 9} - \sqrt{5x + 9}}{h} \][/tex]
Step 4: Rationalize the numerator by multiplying and dividing by the conjugate of the numerator:
[tex]\[ \frac{\sqrt{5x + 5h + 9} - \sqrt{5x + 9}}{h} \cdot \frac{\sqrt{5x + 5h + 9} + \sqrt{5x + 9}}{\sqrt{5x + 5h + 9} + \sqrt{5x + 9}} = \lim_{h \to 0} \frac{(5x + 5h + 9) - (5x + 9)}{h (\sqrt{5x + 5h + 9} + \sqrt{5x + 9})}\][/tex]
Step 5: Simplify the numerator:
[tex]\[ = \lim_{h \to 0} \frac{5h}{h (\sqrt{5x + 5h + 9} + \sqrt{5x + 9})}\][/tex]
Step 6: Cancel out [tex]\( h \)[/tex] in the numerator and denominator:
[tex]\[ = \lim_{h \to 0} \frac{5}{\sqrt{5x + 5h + 9} + \sqrt{5x + 9}}\][/tex]
Step 7: As [tex]\( h \)[/tex] approaches 0, [tex]\( \sqrt{5x + 5h + 9} \)[/tex] approaches [tex]\( \sqrt{5x + 9} \)[/tex]:
[tex]\[ = \frac{5}{\sqrt{5x + 9} + \sqrt{5x + 9}} = \frac{5}{2\sqrt{5x + 9}}\][/tex]
### Final Answer
So, the derivative of the function [tex]\( f(x) = \sqrt{5x + 9} \)[/tex], expressed as the limit, is:
[tex]\[ f'(x) = \frac{5}{2\sqrt{5x + 9}} \][/tex]
This completes the step-by-step derivation using the definition of the derivative.
