Question

F(x)=4x^4+2x^4-x^5 f’(1)-?

Answer 1

Ответ:

$f'(x) = 4 \times 4 {x}^{3} + 2 \times 4 {x}^{3} - 5 {x}^{4} = \\ = 16 {x}^{3} + 8 {x}^{3} - 5 {x}^{4} = 24 {x}^{3} - 5 {x}^{4}$

$f'(1) = 24 - 5 = 19$

Answer 2

To find the derivative of the function $f(x) = 4x^4 + 2x^4 - x^5$, and then evaluate it at $x = 1$, follow these steps:

1. Find the derivative $f'(x)$ of the function $f(x)$ using the power rule for differentiation. The power rule states that if $f(x) = ax^n$, where $a$ and $n$ are constants, then $f'(x) = anx^{n-1}$.

So, for $f(x) = 4x^4 + 2x^4 - x^5$:

$f'(x) = 4 * 4x^{4-1} + 2 * 4x^{4-1} - 1 * 5x^{5-1}$ $f'(x) = 16x^3 + 8x^3 - 5x^4$

2. Now, you want to evaluate $f'(x)$ at $x = 1$. Simply substitute $x = 1$ into the expression for $f'(x)$:

$f'(1) = 16(1)^3 + 8(1)^3 - 5(1)^4$

Simplify the expression:

$f'(1) = 16(1) + 8(1) - 5(1) = 16 + 8 - 5 = 24 - 5 = 19$

So, $f'(1) = 19$.