Question

F(x)=log5(x^2-5x+7); f'(2)=?

Answer 1

Ответ:         f '( 2 ) =  - 1/ ln 5 .

Пошаговое объяснение:

  f(x) = log ₅ ( x² - 5x + 7 );    f'(2) - ?  

  f '( x ) = [ log ₅ ( x² - 5x + 7 ) ]' =  1 * ( x² - 5x + 7 )'/( x² - 5x + 7 ) ln 5 =

            = ( 2x - 5 )/( x² - 5x + 7 ) ln 5 ;        f '( x ) = ( 2x - 5 )/( x² - 5x + 7 ) ln 5 ;

     f '( 2 ) = ( 2*2 - 5 )/( 2² - 5*2 + 7 ) ln 5 = - 1/ 1* ln 5 = - 1/ ln 5 .

Answer 2

To find the derivative of the function $f(x) = \log_5(x^2 - 5x + 7)$ and evaluate it at $x = 2$, you can use the chain rule and the properties of logarithms.

The chain rule states that if you have a composition of functions, such as $f(g(x))$, then the derivative is given by $f'(g(x)) \cdot g'(x)$.

In this case, $f(x) = \log_5(x^2 - 5x + 7)$, and you want to find $f'(2)$, which means you need to find the derivative of $f(x)$ with respect to $x$ and then evaluate it at $x = 2$.

First, let's find $f'(x)$:

$f(x) = \log_5(x^2 - 5x + 7)$

Using the chain rule:

$f'(x) = \frac{1}{\ln(5)} \cdot \frac{1}{x^2 - 5x + 7} \cdot (2x - 5)$

Now, you want to find $f'(2)$:

$f'(2) = \frac{1}{\ln(5)} \cdot \frac{1}{2^2 - 5 \cdot 2 + 7} \cdot (2 \cdot 2 - 5)$

Simplify the expression:

$f'(2) = \frac{1}{\ln(5)} \cdot \frac{1}{4 - 10 + 7} \cdot (4 - 5)$

$f'(2) = \frac{1}{\ln(5)} \cdot \frac{1}{1} \cdot (-1)$

$f'(2) = -\frac{1}{\ln(5)}$

So, $f'(2) = -\frac{1}{\ln(5)}$.