Question

if 2y=5x^2+3;find dy÷dx at x=-2​

Answer 1

Answer:

$2y = 5 {x}^{2} + 3 \\ \\ = > 2\frac{dy}{dx} = 10x \\ \\ = > \frac{dy}{dx} = 5x \\ \\ at \: \: x = - 2 \\ \\ = > \frac{dy}{dx} = - 10$

Answer 2

The value of $\dfrac{dy}{dx}$ = - 10

Step-by-step explanation:

We have,

2y = 5$x^2$ + 3

To find, $\dfrac{dy}{dx}$  (x = - 2​) = ?

∴ 2y = 5$x^2$ + 3

Differentiating both sides w.r.t. x, we get

$\dfrac{d(2y)}{dx}= \dfrac{d(5x^2+3)}{dx}$

⇒ $2\dfrac{dy}{dx}= 5(2x)+0$

⇒ $2\dfrac{dy}{dx}= 10x$

⇒ $\dfrac{dy}{dx}=\dfrac{10x}{2}=5x$

∴ $\dfrac{dy}{dx}=5x$

Put x = - 2, we get

$\dfrac{dy}{dx}$ = 5( -2) = - 10

Thus, the value of $\dfrac{dy}{dx}$ = - 10