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Second derivative of sin(2x) via the chain rule

Let f(x)=sin⁡(2x)f(x) = \sin(2x).

Find f′′(x)f''(x) for f(x)=sin⁡(2x)f(x) = \sin(2x).

Hint

First find f′(x)f'(x) using the chain rule with outer function sin⁡(u)\sin(u) and inner function u=2xu = 2x. Then differentiate f′(x)f'(x) again using the chain rule to get f′′(x)f''(x).

Step-by-step solution

Using the chain rule, f′(x)=cos⁡(2x)⋅ddx[2x]=2cos⁡(2x)f'(x) = \cos(2x)\cdot \frac{d}{dx}[2x] = 2\cos(2x). This is the first derivative, obtained by multiplying the derivative of the outer function evaluated at the inner function by the derivative of the inner function.

Differentiate f′(x)=2cos⁡(2x)f'(x) = 2\cos(2x) again using the chain rule: f′′(x)=2⋅(−sin⁡(2x))⋅2=−4sin⁡(2x)f''(x) = 2\cdot(-\sin(2x))\cdot 2 = -4\sin(2x). So f′′(x)=−4sin⁡(2x)f''(x) = -4\sin(2x).

Third derivative of sin(2x) via the chain rule

Find f′′′(x)f'''(x) for f(x)=sin⁡(2x)f(x) = \sin(2x).

Hint

Differentiate f′′(x)=−4sin⁡(2x)f''(x) = -4\sin(2x) one more time using the chain rule, treating 2x2x as the inner function.

Step-by-step solution

Starting from f′′(x)=−4sin⁡(2x)f''(x) = -4\sin(2x), apply the chain rule: differentiate −4sin⁡(2x)-4\sin(2x) by taking the derivative of −4sin⁡(u)-4\sin(u) with respect to uu, then multiplying by the derivative of u=2xu=2x.

This gives f′′′(x)=−4cos⁡(2x)⋅2=−8cos⁡(2x)f'''(x) = -4\cos(2x)\cdot 2 = -8\cos(2x). So f′′′(x)=−8cos⁡(2x)f'''(x) = -8\cos(2x).

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