Question

Y= sin^3x+sin3x
Find the differentiation

Answer 1

Answer:

given y = sin^3x + sin3x

dy/dx = d(sin^3x)/dx + d(sin3x)/DC

= 3 sin^2x * cos x + cos3x * 3

= 3(sin^2 x cos x + cos 3x)

Answer 2

$</p><p> \sf \rightarrow \quad y = {sin}^{3} \: x + sin \: 3x \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \frac{d}{dx} \bigg( {sin}^{3} \: x + sin \: 3x \bigg) \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \frac{d {(sin \: x)}^{3} }{dx} + \frac{d(sin \: 3x)}{dx} \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \bigg\{ \frac{d {(sin \: x)}^{3} }{d \: sin \: x} \: \cdot \: \frac{d ( sin \: x)}{dx} \bigg\} + \bigg\{ \frac{d(sin \: 3x)}{d \: 3x} \: \cdot \: 3\frac{dx}{dx} \bigg\} \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = \bigg \{ 3 {sin}^{2} \: x \cdot \: cos \: x \bigg \} - \bigg \{ cos \: 3x \: \cdot \: 3 \bigg\} \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = 3 {sin}^{2} \: x \: cos \: x - 3cos \: 3x \\ \\ \sf \rightarrow \quad \frac{dy}{dx} = 3 ( sin^{2} \: x \: cos \: x - cos \: 3x )</p><p>$