Question

y = (x + cos x)sinx solve with log differentiation

Answer 1

\[\begin{align*}y&=(x+\cos x)\sin x\\
\ln y&=\ln((x+\cos x)\sin x)\\
\frac{d}{dx}\left[\ln y\right]&=\frac{d}{dx}\left[\ln((x+\cos x)\sin x)\right]\\
\frac{1}{y}\frac{dy}{dx}&=\frac{d}{dx}\left[\ln(x+\cos x)+\ln\sin x\right]\\
\frac{1}{y}\frac{dy}{dx}&=\frac{1-\sin x}{x+\cos x}+\frac{\cos x}{\sin x}\\
\frac{dy}{dx}&=y\left(\frac{1-\sin x}{x+\cos x}+\frac{\cos x}{\sin x}\right)\\
\frac{dy}{dx}&=(x+\cos x)\sin x\cdot\left(\frac{1-\sin x}{x+\cos x}+\frac{\cos x}{\sin x}\right)\\
\frac{dy}{dx}&=\sin x(1-\sin x)+(x+\cos x)\cos x\\
\frac{dy}{dx}&=\sin x-\sin^2 x+x\cos x+\cos^2 x\\
\frac{dy}{dx}&=\sin x+x\cos x+\cos2 x\\

\end{align*}\]