# How do you use the definition of a derivative to find the derivative of #f(x) = sqrtx + 2# to calculate f'(2)?

By definition of derivative:

simplify:

By signing up, you agree to our Terms of Service and Privacy Policy

To find the derivative of ( f(x) = \sqrt{x} + 2 ), we use the definition of a derivative, which states:

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

Plugging in ( f(x) = \sqrt{x} + 2 ), we have:

[ f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} + 2 - (\sqrt{x} + 2)}{h} ]

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

To simplify this expression, we use the conjugate pair:

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

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

[ f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} ]

Now, we plug in ( x = 2 ) to find ( f'(2) ):

[ f'(2) = \frac{1}{\sqrt{2+h} + \sqrt{2}} ]

Taking the limit as ( h \to 0 ):

[ f'(2) = \frac{1}{\sqrt{2} + \sqrt{2}} = \frac{1}{2\sqrt{2}} ]

By signing up, you agree to our Terms of Service and Privacy Policy

- What is the equation of the line tangent to # f(x)=sqrt(e^x-x) # at # x=1#?
- How do you find the tangent line of #f(x) = 3-2x^2 # at x=-1?
- What is the derivative of #6 (3^(2x-1))#?
- How do you find the equation of the tangent line to the graph of #f(x)=sqrt(x-1)# at point #(5,2)#?
- How do you find the equation of the line tangent to #y=cosx# at x=π/4?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7