Question

Y=(6x^4-3x^2)(x^3+7x)y'=?

Answer 1

Ответ:

Объяснение:

y = 6x^7 + 42x^5 - 3x^5 - 21x^3 = 6x^7 + 39x^5 - 21x^3

y' = 42x^6 + 195x^4 - 63x^2

Answer 2

To find the derivative of the function Y with respect to x, which is $Y = (6x^4 - 3x^2)(x^3 + 7x)$, we will use the product rule. The product rule states that if you have two functions $u(x)$ and $v(x)$, the derivative of their product is given by:

$(uv)' = u'v + uv'$

In this case, let $u(x) = 6x^4 - 3x^2$ and $v(x) = x^3 + 7x$. We need to find the derivatives of $u(x)$ and $v(x)$ with respect to $x$, and then apply the product rule.

Let's start by finding the derivatives:

$u'(x) = \frac{d}{dx}(6x^4 - 3x^2) = 24x^3 - 6x$

$v'(x) = \frac{d}{dx}(x^3 + 7x) = 3x^2 + 7$

Now, applying the product rule:

$Y' = (6x^4 - 3x^2)(3x^2 + 7) + (24x^3 - 6x)(x^3 + 7x)$ $Y' = (18x^6 + 42x^4 - 9x^2 - 21x^2) + (24x^6 + 168x^4 - 6x^4 - 42x^2)$ $Y' = 42x^6 + 210x^4 - 51x^2$

So, the derivative of $Y$ with respect to $x$ is $Y' = 42x^6 + 210x^4 - 51x^2$.