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2021-10-19

If $f\left(x\right)={\mathrm{log}}_{a}x$ , show that $\frac{f(x+h)-f\left(x\right)}{h}={\mathrm{log}}_{a}{(1+\frac{h}{x})}^{\frac{1}{h}},h\ne c0$

Arham Warner

Skilled2021-10-20Added 102 answers

Step 1

The difference quotient$\frac{f(x+h)-f\left(x\right)}{h}$ can be used to calculate the first derivative of the function by taking the limit $h\to 0$ . It will give f'(x) if it exists.

For the given problem, two properties of logarithms will be used. The first is the difference property, which is$\mathrm{log}}_{a}x-{\mathrm{log}}_{a}y={\mathrm{log}}_{a}\frac{x}{y$ . The second one is the exponent property of logarithms, which is $y{\mathrm{log}}_{a}x={\mathrm{log}}_{a}{x}^{y}$

Step 2

Substitute$f\left(x\right)={\mathrm{log}}_{a}x$ in the difference quotient $\frac{f(x+h)-f\left(x\right)}{h}$ and simplify using the properties of logarithms.

$\frac{f(x+h)-f\left(x\right)}{h}=\frac{{\mathrm{log}}_{a}(x+h)-{\mathrm{log}}_{a}x}{h}=$

$=\frac{{\mathrm{log}}_{a}\frac{x+h}{x}}{h}$

$=\frac{1}{h}{\mathrm{log}}_{a}(1+\frac{h}{x})$

$={\mathrm{log}}_{a}{(1+\frac{h}{x})}^{\frac{1}{h}}$

It is defined for$h\ne 0\text{}\text{as}\text{}\frac{1}{h}$ is not defined.

The difference quotient

For the given problem, two properties of logarithms will be used. The first is the difference property, which is

Step 2

Substitute

It is defined for

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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