Differentiate (x-3) (x^2+5) with respect to x
use the product rule
y' = (1)(x^2+5) + (x-3)(2x) = 3x^2 - 6x+ 5
OR
( x - 3 ) ( x² + 5 ) = x² ∙ x - x² ∙ 3 + 5 ∙ x - 5 ∙ 3
( x - 3 ) ( x² + 5 ) = x³ - 3 x² + 5 x - 15
d / dx ( x³ - 3 x² + 5 x - 15 ) = 3 x² - 3 ∙ 2 x + 5
d / dx ( x³ - 3 x² + 5 x - 15 ) = 3 x² - 6 x + 5
To differentiate the expression (x-3)(x^2+5) with respect to x, we can use the product rule of differentiation. The product rule states that the derivative of a product of two functions u(x) and v(x) is given by:
(d/dx)(u(x) * v(x)) = u'(x) * v(x) + u(x) * v'(x)
So, let's differentiate the expression step by step:
First, let u(x) = (x-3) and v(x) = (x^2+5).
The derivative of u(x) with respect to x, denoted as u'(x), is 1, since the derivative of x with respect to x is 1, and the derivative of a constant (like -3) is 0.
The derivative of v(x) with respect to x, denoted as v'(x), is 2x, since we apply the power rule of differentiation, which states that the derivative of x^n with respect to x is n*x^(n-1), where n is any real number.
Now, we can apply the product rule:
(d/dx)((x-3)(x^2+5)) = u'(x) * v(x) + u(x) * v'(x)
= 1 * (x^2+5) + (x-3) * (2x)
= x^2 + 5 + 2x(x-3)
= x^2 + 5 + 2x^2 - 6x
= 3x^2 - 6x + 5
Therefore, the derivative of (x-3)(x^2+5) with respect to x is 3x^2 - 6x + 5.