Question

F(x)=4x+3/4x+1;f'(x)-?

Answer 1

Ответ:4-3/(4x^2)

Пошаговое объяснение:

От 4х =4, а от 3/(4х) производная будет 3/4*(-1/x^2).

Answer 2

Ответ:

напишем производную частного

((4x+3)'(4x+1)-(4x+3)(4x+1)')/(4x+1)^2=4(4x+1)-4(4x+3)/(4x+1)^2=-8/(4x+1)^2

Answer 3

To find the derivative of the function $f(x) = \frac{4x+3}{4x+1}$, we'll use the quotient rule. The quotient rule states that if we have a function $f(x) = \frac{u(x)}{v(x)}$, then its derivative $f'(x)$ is given by:

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

where $u'(x)$ and $v'(x)$ are the derivatives of $u(x)$ and $v(x)$ respectively.

For the given function $f(x) = \frac{4x+3}{4x+1}$, we can identify $u(x) = 4x+3$ and $v(x) = 4x+1$. Let's find their derivatives first:

$u'(x) = \frac{d}{dx}(4x+3) = 4$ (derivative of $4x$ is $4$, derivative of a constant term $3$ is $0$),

$v'(x) = \frac{d}{dx}(4x+1) = 4$ (derivative of $4x$ is $4$, derivative of a constant term $1$ is $0$).

Now, we can apply the quotient rule:

$f'(x) = \frac{(4)(4x+1) - (4x+3)(4)}{(4x+1)^2}$

Now, simplify the expression:

$f'(x) = \frac{16x + 4 - 16x - 12}{(4x+1)^2}$

$f'(x) = \frac{-8}{(4x+1)^2}$

So, the derivative of the function $f(x) = \frac{4x+3}{4x+1}$ is $f'(x) = \frac{-8}{(4x+1)^2}$.