Question

second order derivative of sin 3x-2​

Answer 1

Answer:

-6 is your correct answer it helps you

Answer 2

Answer:

The second-order derivative is -9sin(3x-2).

Step-by-step explanation:

Given:

sin(3x-2)

To find:

Second-order derivative

Solution:

Let y = sin(3x-2)

Differentiating w.r.t x, we get

$\frac{dy}{dx} = \frac{d[sin(3x-2)]}{dx}$

We know that the derivative of sin(x) is cos(x). Using this and applying the chain rule, we get

$\frac{dy}{dx} = cos(3x-2)\frac{d(3x-2)}{dx}$

$\frac{dy}{dx} = cos(3x-2)[\frac{d(3x)}{dx}-\frac{d(2)}{dx}]$

The derivative of x is 1 and that of a constant is 0. Therefore,

$\frac{dy}{dx} = cos(3x-2)[3(1) - 0]$

$\frac{dy}{dx} = 3cos(3x-2)$

This is the first derivative of sin(3x-2) w.r.t x.

To find the second derivative, we again differentiate dy/dx w.r.t. x. Hence,

$\frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx})$

Substituting the obtained value of dy/dx, we get

$\frac{d^2y}{dx^2} = \frac{d[3cos(3x-2)]}{dx}$

$\frac{d^2y}{dx^2} = 3\frac{d[cos(3x-2)]}{dx}$

The derivative of cos(x) is -sin(x). Using this and the chain rule, we get

$\frac{d^2y}{dx^2} = -3sin(3x-2)\frac{d(3x-2)}{dx}$

$\frac{d^2y}{dx^2} = -3sin(3x-2)[\frac{d(3x)}{dx} - \frac{d(2)}{dx}]$

$\frac{d^2y}{dx^2} = -3sin(3x-2)[3-0]$

$\frac{d^2y}{dx^2} = -9sin(3x-2)$

Therefore, the second derivative of sin(3x-2) is -9sin(3x-2).

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