$$\text { Find } \frac { d y } { d x }$$
-$$e ^ { 6 y } = \cos ( 8 x + y )$$
A) $\frac { - 8 \sin ( 8 x + y ) } { 6 e ^ { 6 y } + \sin ( 8 x + y ) }$
B) $\ln \cos ( 8 x + y )$
C) $\frac { \sin ( 8 x + y ) } { 6 e ^ { 6 y } }$
D) $\frac { - 8 \sin ( 8 x + y ) + 8 } { 6 e ^ { 6 y } - 1 }$

Question

Quizplus · Accepted Answer

Using the chain rule, we get:
\begin{align*}
\frac{d}{dx}(e^{6y}) &= \frac{d}{dx}(\cos(8x+y)) \
6e^{6y}\frac{dy}{dx} &= -8\sin(8x+y) + \frac{dy}{dx} \
\frac{dy}{dx}(1- 6e^{6y}) &= -8\sin(8x+y) \
\frac{dy}{dx} &= \frac{-8\sin(8x+y)}{6e^{6y}+\sin(8x+y)}
\end{align*}
Therefore, the correct answer is (A).

Find dydx\text { Find } \frac { d y } { d x } Find dxdy​

$\text { Find } \frac { d y } { d x }$