frac{d}{dx} [3 sin(60^{circ} - x^{circ}) - 4 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) First, convert the degree measure to radians: $x^{\circ} = \frac{\pi x}{180}$.We use the identity $\cos(30^{\circ} + x^{\circ}) = \sin(90^{\circ}

Vedclass · Accepted Answer

$\frac{\pi}{60} \cos(3x^{\circ})$

Vedclass · Answer

(C) First, convert the degree measure to radians: $x^{\circ} = \frac{\pi x}{180}$.We use the identity $\cos(30^{\circ} + x^{\circ}) = \sin(90^{\circ} - (30^{\circ} + x^{\circ})) = \sin(60^{\circ} - x^{\circ})$.Let $A = 60^{\circ} - x^{\circ}$. The expression becomes $3 \sin A - 4 \sin^3 A$.Using the trigonometric identity $3 \sin A - 4 \sin^3 A = \sin(3A)$, we get $\sin(3(60^{\circ} - x^{\circ})) = \sin(180^{\circ} - 3x^{\circ}) = \sin(3x^{\circ})$.Now, express this in terms of radians: $f(x) = \sin(3 \cdot \frac{\pi x}{180}) = \sin(\frac{\pi x}{60})$.Differentiating with respect to $x$ using the chain rule: $\frac{d}{dx} [\sin(\frac{\pi x}{60})] = \cos(\frac{\pi x}{60}) \cdot \frac{\pi}{60} = \frac{\pi}{60} \cos(3x^{\circ})$.

$\frac{d}{dx} [3 \sin(60^{\circ} - x^{\circ}) - 4 \cos^3(30^{\circ} + x^{\circ})] = \rule{1cm}{0.15mm}$

Similar Questions

$\frac{d}{dx} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) = $ . . . . . .

If $f(x) = |x^2 - 3x + 2|$,then $\frac{df}{dx} = $

If $3 f(x)-2 f\left(\frac{1}{x}\right)=x$,then $f^{\prime}(2)=$

Find the derivative: $\frac{d}{dx}(x e^{x^2}) = $

Find the derivative of the function: $\frac{x}{1+\tan x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module