Question

Find the derivative of f(x) = 7/x at x = 1.

Answer 1

HINT: \(\dfrac{7}{x} = 7x^{-1}\)

You can apply power rule.

Answer 2

What's the power rule? I've only learned about the formula [f(x + h) - f(x)]/h

Answer 3

wow, you teacher is cruel, then.

Give me time to figure that out.

Answer 4

OK, so while taking derivative, we have \(f'(x) = \displaystyle\lim_{h\to0}\dfrac{\dfrac{7}{x+h} - \dfrac{7}{x}}{h}\)

In numerator, you can find GCD, which is \(x(x+h)\), right?

So you have \(f'(x) = \displaystyle\lim_{h\to0}\dfrac{\dfrac{7}{x+h}\cdot\dfrac{x}{x} - \dfrac{7}{x}\cdot\dfrac{x+h}{x+h}}{h} =\lim_{h\to0} \dfrac{\dfrac{7x-7(x+h)}{x(x+h)}}{h}\)

\[=\lim_{h\to0}\dfrac{\dfrac{7x-7x-7h}{x(x+h)}}{h}=\lim_{h\to0}\dfrac{\dfrac{-7h}{x(x+h)}}{h}=\lim_{h\to0}\dfrac{-7h}{xh(x+h)} \\~\\ =\lim_{h\to0}\dfrac{-7\cancel h}{x\cancel h(x+h)} = \lim_{h\to0}\dfrac{-7}{x(x+h)} = \lim_{h\to0}\dfrac{-7}{x^2+xh}\]

Now you can do direct substitution; \(\cdots = \dfrac{-7}{x^2+x(0)} = \dfrac{-7}{x^2}\)

Now plug in 1 for x.

Answer 5

You will learn about power rule soon... it will save you a lot of works...

Answer 6

@EuroSpeeder

Answer 7

Oh wow that makes so much sense! So:
-7/(1)^2

-7/1 = -7?

Answer 8

@geerky42

Answer 9

Right

Answer 10

Thank you so much!

Answer 11

no problem