Mathematics: Find the derivative of the function using the definition of derivative. f(x) = 6 + x 1 − 6x f (x) = state the domain of the function. (enter your answer in interval notation.) state the domain of its derivative. (enter your answer in interval notation.)

The answer is 799 ur welcome...

Find the derivative of the function using the definition

dgayles8761

07.01.2022

$f'(x)=\dfrac{37}{(1-6x)^2}$

Domain of f(x): $(-\infty,\frac{1}{6})\cup (\frac{1}{6},\infty)$ .

Domain of f'(x): $Domain=(-\infty,\frac{1}{6})\cup (\frac{1}{6},\infty)$ .

Step-by-step explanation:

Consider the given function

$f(x)=\dfrac{6+x}{1-6x}$

Domain of this function is all real numbers except those numbers for which denominator is equal to 0.

1-6x=0

1=6x

Divide both sides by 6.

$\frac{1}{6}=x$

The function f(x) is not defined for 1/6. So, the domain of the function f(x) is

$Domain=(-\infty,\frac{1}{6})\cup (\frac{1}{6},\infty)$

Find the derivative of the function using the definition of derivative.

$f'(x)=lim_{x\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{x\rightarrow 0}\dfrac{\dfrac{6+(x+h)}{1-6(x+h)}-\dfrac{6+x}{1-6x}}{h}$

$f'(x)=lim_{x\rightarrow 0}\frac{1}{h}\dfrac{(6+x+h)(1-6x)-(6+x)(1-6x-6h)}{(1-6x-6h)(1-6x)}$

$f'(x)=lim_{x\rightarrow 0}\frac{1}{h}\dfrac{(6+x)(1-6x)+h(1-6x)-(6+x)(1-6x)-(6+x)(-6h)}{(1-6x-6h)(1-6x)}$

$f'(x)=lim_{x\rightarrow 0}\frac{h}{h}\dfrac{1-6x+36+6x}{(1-6x-6h)(1-6x)}$

$f'(x)=lim_{x\rightarrow 0}\dfrac{37}{(1-6x-6h)(1-6x)}$

Apply limit.

$f'(x)=\dfrac{37}{(1-6x-6(0))(1-6x)}$

$f'(x)=\dfrac{37}{(1-6x)^2}$

The derivative of given function is $f'(x)=\dfrac{37}{(1-6x)^2}$ .

(1-6x)^2=0

1-6x=0

1=6x

Divide both sides by 6.

$\frac{1}{6}=x$

The function f'(x) is not defined for 1/6. So, the domain of the function f'(x) is

$Domain=(-\infty,\frac{1}{6})\cup (\frac{1}{6},\infty)$ .

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