Find the derivative of x-4 all under the square root sign. Using f(a+ delta x) -f(a) all over delta x.

Question

anonymous · Answer

$$f(a)=\sqrt{a-4}\
f'(a)=\lim_{\Delta x\to0}\frac{f(a+\Delta x)-f(a)}{\Delta x}\
f'(a)=\lim_{\Delta x\to0}\frac{\sqrt{(a+\Delta x) -4}-\sqrt{a-4}}{\Delta x}$$
Multiply the numerator and denominator by the conjugate of the numerator; that is, by
$$\frac{\sqrt{(a+\Delta x) -4}+\sqrt{a-4}}{\sqrt{(a+\Delta x) -4}+\sqrt{a-4}}$$
Then simplify.

anonymous · Answer

Does it come out to be 1 over 2 times the square root of x-4?

anonymous · Answer

Yes, that's right. I used an $a$ instead of $x$ above to follow the definition you're using.

anonymous · Answer

okay! Thank you so much!

anonymous · Answer

You're welcome.