How do you use the limit definition to find the derivative of #y=sqrt(-4x-2)#?

The limit definition of the derivative is given by:

Putting this into the above definition:

Now we attempt to simplify.

We can verify this answer by taking the derivative directly (using the chain rule):

To use the limit definition to find the derivative of ( y = \sqrt{-4x - 2} ), follow these steps:

1. Start with the given function: [ y = \sqrt{-4x - 2} ]

2. Use the definition of the derivative: [ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} ]

3. Substitute the given function into the definition: [ f'(x) = \lim_{h \to 0} \frac{\sqrt{-4(x + h) - 2} - \sqrt{-4x - 2}}{h} ]

4. Rationalize the numerator by multiplying by the conjugate: [ f'(x) = \lim_{h \to 0} \frac{\sqrt{-4(x + h) - 2} - \sqrt{-4x - 2}}{h} \times \frac{\sqrt{-4(x + h) - 2} + \sqrt{-4x - 2}}{\sqrt{-4(x + h) - 2} + \sqrt{-4x - 2}} ]

5. Simplify the numerator: [ f'(x) = \lim_{h \to 0} \frac{-4(x + h) - 2 + 4x + 2}{h(\sqrt{-4(x + h) - 2} + \sqrt{-4x - 2})} ]

6. Combine like terms in the numerator: [ f'(x) = \lim_{h \to 0} \frac{-4h}{h(\sqrt{-4(x + h) - 2} + \sqrt{-4x - 2})} ]

7. Cancel out the common factor of ( h ): [ f'(x) = \lim_{h \to 0} \frac{-4}{\sqrt{-4(x + h) - 2} + \sqrt{-4x - 2}} ]

8. Substitute ( h = 0 ) into the expression: [ f'(x) = \frac{-4}{\sqrt{-4x - 2} + \sqrt{-4x - 2}} ]

9. Simplify the expression: [ f'(x) = \frac{-4}{2\sqrt{-4x - 2}} ]

10. Further simplify the expression by dividing numerator and denominator by 2: [ f'(x) = \frac{-2}{\sqrt{-4x - 2}} ]

Therefore, the derivative of ( y = \sqrt{-4x - 2} ) using the limit definition is ( f'(x) = \frac{-2}{\sqrt{-4x - 2}} ).