Use logarithmic differentiation to find the derivative of y.
-$y = \sin x \sqrt { 5 x + 2 }$
A) $\sin x \sqrt { 5 x + 2 } \left( \frac { 1 } { \sin x \cdot \cos x } + \frac { 5 } { 5 x + 2 } \right)$
B) $\sin x \sqrt { 5 x + 2 } \left( \cot x + \frac { 5 } { 2 ( 5 x + 2 ) } \right)$
C) $\frac { 10 \cot x + 9 \cot x } { 5 x + 4 }$
D) $\sin x \sqrt { 5 x + 2 } \left( \ln \sin x + \frac { 1 } { 2 } \ln ( 5 x + 2 ) \right)$

Question

Quizplus · Accepted Answer

Using logarithmic differentiation, we take the natural logarithm of both sides and differentiate:

\begin{align*}
\ln y &= \ln(\sin x \sqrt{5x+2}) \
\ln y &= \ln (\sin x) + \ln (\sqrt{5x+2}) \
\ln y &= \ln (\sin x) + \frac{1}{2} \ln (5x+2) \
\frac{1}{y} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{1}{\sin x} \cdot \cos x + \frac{1}{2(5x+2)} \cdot 5 \
\frac{\mathrm{d}y}{\mathrm{d}x} &= y \left( \frac{\cos x}{\sin x} + \frac{5}{2(5x+2)} ight) \
\frac{\mathrm{d}y}{\mathrm{d}x} &= \sin x \sqrt{5x+2} \left( \cot x + \frac{5}{2(5x+2)} ight)
\end{align*}

Therefore, the answer is (B).

Use Logarithmic Differentiation to Find the Derivative of Y y=sin⁡x5x+2y = \sin x \sqrt { 5 x + 2 }y=sinx5x+2​

Use Logarithmic Differentiation to Find the Derivative of Y $y = \sin x \sqrt { 5 x + 2 }$