Question

Differentiate w.r.t.'x'
y=(x+3)(x^(2)-4)

Answer 1

Answer:

$\frac{dy}{dx}=3x^2+6x-4$

Step-by-step explanation:

$y =(x+3)(x^2-4)$

$y=x^3+3x^2-4x-12$

Differentiating with respect to x, we get

$\frac{dy}{dx} = 3x^{3-1} + 2 \times3 x^{2-1}-4\times 1- 0\\\frac{dy}{dx}=3x^2+6x-4$

Answer 2

Given:

$y=(x+3)(x^2-4)$

To Find:

Differentiation of $y=(x+3)(x^2-4)$ = ?

Solution:

On solving brackets, we get

⇒  $y=x^3-4x+3x^2-12$

⇒  $y=x^3+3x^2-4x-12$

On differentiating the above equation with respect to "x", we get

⇒  $\frac{dy}{dx}=3x^2+6x-4$