Question

Find the derivatives of the following function from the definition in terms of x, f(x) and f′(x).
g(x) = f(x)^2

Answer 1

By the chain-power rule, you have
\[\dfrac{d}{dx}g(x) = 2 f(x) \cdot f'(x)\]

Answer 2

i cant use the chain rule... I have to use the definition

Answer 3

Oh, sorry! I didn't catch that detail.
By the definition, you have
\[g'(x)=\displaystyle \lim_{h\to 0} \dfrac{g(x+h)-g(x)}{h}\]
Since\[g(x)=(f(x))^2\] the limit becomes
\[g'(x)=\displaystyle \lim_{h\to 0} \dfrac{[f(x+h)]^2-[f(x)]^2}{h}\]
This might not be the obvious step to make, but this is what you do next:
\[g'(x)=\displaystyle \lim_{h\to 0} \dfrac{[f(x+h)]^2-[f(x)]^2+2f(x)f(x+h)-2f(x)f(x+h)+[f(x)]^2-[f(x)]^2}{h}\]
You should be able to tell that I haven't fundamentally changed the expression in the numerator, since (expression) - (expression) = 0, and (something) + 0 = (something).
Next, you can rearrange the terms and separate the limit into two:
\[\displaystyle \lim_{h\to 0} \dfrac{[f(x+h)]^2-2f(x)f(x+h)+[f(x)]^2}{h} + \lim_{h\to 0} \dfrac{2f(x)f(x+h)-2[f(x)]^2}{h}\]
Note that the numerator in the first limit can be factored easily. In the second limit, factor out a 2f(x):
\[\displaystyle \lim_{h\to 0} \dfrac{[f(x+h) - f(x)]^2}{h} + 2f(x)\lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h}\]
\[\displaystyle \left[ \lim_{h\to 0} \dfrac{f(x+h) - f(x)}{h}\right]\lim_{h\to 0} [f(x+h)-f(x)] + 2f(x)\cdot f'(x)\]
\[\displaystyle f'(x)\lim_{h\to 0} [f(x+h)-f(x)] + 2f(x)\cdot f'(x)\]
\[\displaystyle f'(x)\cdot [f(x)-f(x)] + 2f(x)\cdot f'(x)\]
\[\displaystyle f'(x)\cdot 0 + 2f(x)\cdot f'(x)\]
\[2f(x) \cdot f'(x)\]

Answer 4

The cut-off text is just
\[]^2\]

Answer 5

ok wow.. I would not have figured out like any of that... can u help me on a few other ones too?

Answer 6

Sure. The strategy I used was finding the derivative first using the chain/product rules, and then working backward from that.

Answer 7

yeah. I'm havent learned the chain rule yet, and just not allowed to use any of the other rules for these yet.
\[g(x)= f(x)^{3}\]
\[g(x)= \sqrt{f(x)}\]
\[g(x)=\sqrt[3]{f(x)}\]

Answer 8

May I ask what level calc course this is for?

Answer 9

calc 1

Answer 10

Well, finding the first function's derivative uses similar (but some more intricate) reasoning as for the earlier one you asked, and the other two are too difficult for me (or I'm just not seeing the right method right away). This might take some time...

Answer 11

lol it's fine... I don't eve know what to do so

Answer 12

Alright, for
\[g(x)=\sqrt{f(x)}\]the derivative would be
\[\displaystyle \lim_{h\to 0} \dfrac{\sqrt{f(x+h)}-\sqrt{f(x)}}{h}\]
Multiply the numerator and denominator by the conjugate of the numerator:
\[\dfrac{\sqrt{f(x+h)}+\sqrt{f(x)}}{\sqrt{f(x+h)}+\sqrt{f(x)}}\]
Then, you have
\[\displaystyle \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h(\sqrt{f(x+h)}+\sqrt{f(x)})}\]
\[\displaystyle \lim_{h\to 0} \dfrac{f(x+h)-f(x)}{h} \lim_{h\to 0} \dfrac{1}{\sqrt{f(x+h)}+\sqrt{f(x)}}\]
\[f'(x) \lim_{h\to 0} \dfrac{1}{\sqrt{f(x+h)}+\sqrt{f(x)}}\]
\[f'(x) \dfrac{1}{\sqrt{f(x)}+\sqrt{f(x)}}\]
\[f'(x) \dfrac{1}{2\sqrt{f(x)}}\]
I'll see if I can get any progress on the last two.