Question

Find the derivative of f(x) = negative 9 divided by x at x = 6.

Answer 1

First, find f(x+h) = (-9/(x+h)) ... just substitue (x+h) for every x.

Then plug in f(x+h) and f(x) in the difference quotient formula.

Lim/h->0\ [f(x+h) - f(x)]/h

Or, [(-9/ (x+h)) - -9/x] /h
Or, [(-9/ (x+h)) + 9/x]/h
Or, [(-9x + 9x + 9h)/ x(x+h)] / h
Or, [9h / (x^2 + hx)]/h .............................................. add like terms
Or, ( 9h / (x^2 + hx)) * h/1 .................divide by 7 is equivalent to multiply by 1/7
Or, 9/ (x^2 + hx) .................................cancle out the h
Or, 9/ x^2 ........................................... plug in 0 for h

Now, the function asks for x=6, so we plug in 6 for every x
so, the limit = 9/ 6^2 = 9/36 = 1/4.

Answer 2

The derivative of $f(x) = \frac{9}{x}$ evaluated at $x = 6$ is $-\frac{1}{6}$.

In this question, we will use the Derivative Rules and the direct Evaluation of the resulting Function at a given value of $x$. Let $f(x) = \frac{9}{x}$, which is equivalent to $f(x) = 9\cdot x^{-1}$, so that the following Derivative Rule:

$\frac{d}{dt} (x^{n}) = n\cdot x^{n-1}$, $n \ne 0$ (1)

Hence, we find that the derivative of $f(x)$ is:

$f'(x) = -9\cdot x^{-2}$

$f'(x) = -\frac{9}{x^{2}}$

If we know that $x = 6$, then the derivative of the function is:

$f'(6) = -\frac{9}{6^{2}}$

$f'(6) = -\frac{9}{36}$

$f'(6) = -\frac{1}{4}$

The derivative of $f(x) = \frac{9}{x}$ evaluated at $x = 6$ is $-\frac{1}{6}$.

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