Question

y = cos^-1 (3cosx-2sinx/root13) find dy/dx

Answer 1

Step-by-step explanation:

$y = {cos}^{ - 1} ( \: \dfrac{3cosx - 2sinx}{ \sqrt{13} } ) \\ = {cos}^{ - 1} ( \: \dfrac{3}{ \sqrt{13} } \: \: cosx - \dfrac{2}{ \sqrt{13} } \: \: sinx \: ) \\ let \\ \dfrac{3}{ \sqrt{13} } = cos \alpha \: \: and \: \: sin \alpha = \dfrac{2}{ \sqrt{13} } \\ so \\ tan \alpha = \dfrac{2}{3} \\ \alpha = {tan}^{ - 1} ( \dfrac{2}{3} ) \\ now \\ y = {cos}^{ - 1} (cosx \: \: cos \alpha - \: sinx \: sin \alpha ) \\ y = {cos}^{ - 1} \: \: cos \: (x + \alpha ) \\ y = x + \alpha \\ y = x + {tan}^{ - 1} ( \dfrac{2}{3} ) \\ differentiating \: with \: respect \: to \: x \\ \dfrac{dy}{dx} = \dfrac{d}{dx} (x) + \dfrac{d}{dx} ( {tan}^{ - 1} ( \frac{2}{3} ) \: ) \\ \dfrac{dy}{dx} = 1 + 0 = 1$