Question

Using the difference quotient, find f'(x) from 2sqrtx

Answer 1

I know the answer is \[\frac{ 1 }{ \sqrt{x} }\]

Answer 2

But I get only as far as \[\frac{ 2x + 2h + 2x}{ 2\sqrt{x} + 2\sqrt{x}}\]

Answer 3

ok so difference quotient: \(\dfrac{f(x+h)-f(x)}{h}\)

I would find what \(f(x+h)\) is and state what \(f(x)\) is separately, then just plug it into the formula. \[f(x) = 2\sqrt{x}\]\[f(x+h) = 2\sqrt{x+h}\]\[\begin{align} f'(x)= \dfrac{2\sqrt{x+h}-2\sqrt{x}}{h} \\ &=\frac{2\sqrt{x+h} -2\sqrt{x}}{h} \cdot \frac{2\sqrt{x+h}+2\sqrt{x}}{2\sqrt{x+h}+2\sqrt{x}} \\ &=\frac{4(x+h)-4x}{h(2\sqrt{x+h}+2\sqrt{x})} \\&=\frac{4h}{h(2\sqrt{x+h}+2\sqrt{x})} \\ &=\color{red}{\frac{4}{2\sqrt{x+h}+2\sqrt{x}}} \end{align} \]

Answer 4

THANK YOU

Answer 5

did you follow?

Answer 6

Yes I messed up
I saw my mistake
you da bestest!

Answer 7

Remember that: \((a-b)(a+b) = a^2-b^2\)

Answer 8

thanks i follow....:)

Answer 9

Awesome

Answer 10

i love u