$\text { Find } \frac { d y } { d x } \text { by implicit differentiation given that } \tan ( 4 x + y ) = 4 x \text {. Use the original }$ equation to simplify your answer.
A) $\frac { d y } { d x } = \frac { 4 x } { x ^ { 2 } + 1 }$
B) $\frac { d y } { d x } = - \frac { 4 x ^ { 2 } } { x ^ { 2 } - 1 }$
C) $\frac { d y } { d x } = \frac { 4 x ^ { 2 } } { x ^ { 2 } + 1 }$
D) $\frac { d y } { d x } = - \frac { 4 x ^ { 2 } } { x ^ { 2 } + 1 }$
E) $\frac { d y } { d x } = \frac { 4 x ^ { 2 } } { x ^ { 2 } - 1 }$

Question

Quizplus · Accepted Answer

Taking the derivative of the given equation with respect to $x$ using chain rule, we get:

\begin{align*}
\sec^2(4x+y) \cdot \frac{d}{dx}(4x+y) &= 4 \
\sec^2(4x+y) \cdot (1+\frac{dy}{dx}) &= 4 \
1+\frac{dy}{dx} &= \frac{4}{\sec^2(4x+y)} \
\end{align*}

Using the trigonometric identity $\sec^2(x) = 1+	an^2(x)$, we can rewrite the denominator as follows:

\begin{align*}
\sec^2(4x+y) &= 1+	an^2(4x+y) \
&= 1+(4x)^2+y^2+2(4x)(y)'
\end{align*}

Now, solving for $y'$ gives:
\begin{align*}
1+\frac{dy}{dx} &= \frac{4}{1+(4x)^2+y^2+2(4x)(y)'} \
1+\frac{dy}{dx} &= \frac{4}{1+(4x)^2+y^2+8x+2y'} \
1+(4x)^2+y^2+8x+2y' &= \frac{4}{1+\frac{dy}{dx}} \
1+(4x)^2+y^2+8x+2y' &= \frac{4}{\frac{dy}{dx}+1} \ 
\frac{dy}{dx}\left(1+\frac{4}{\frac{dy}{dx}+1}ight) &= -(4x)^2-y^2-8x-1 \
\frac{dy}{dx}\cdot\frac{\frac{dy}{dx}+5}{\frac{dy}{dx}+1} &= -(4x)^2-y^2-8x-1 \
\frac{dy}{dx}+5 &= \frac{-(4x)^2-y^2-8x-1}{\frac{dy}{dx}+1} \
\frac{dy}{dx}\left(\frac{\frac{dy}{dx}+6}{\frac{dy}{dx}+1}ight) &= -(4x)^2-y^2-8x-1 \
\frac{dy}{dx} &= -\frac{(4x)^2+y^2+8x+1}{\frac{dy}{dx}+1} \cdot \frac{\frac{dy}{dx}+1}{\frac{dy}{dx}+6} \
\frac{dy}{dx} &= -\frac{(4x)^2+y^2+8x+1}{(\frac{dy}{dx})^2+7\cdot\frac{dy}{dx}+6} \
\frac{dy}{dx} &= -\frac{(4x)^2+y^2+8x+1}{(\frac{dy}{dx}+1)(\frac{dy}{dx}+6)}
\end{align*}

To simplify this expression, we substitute the given equation $	an(4x+y)=4x$ and solve for $y^2$:

\begin{align*}
	an^2(4x+y) &= (4x)^2 \
\sec^2(4x+y)-1 &= (4x)^2 \
1+	an^2(4x+y)-1 &= (4x)^2 \
y^2 &= (4x)^2-	an^2(4x+y) \
&= (4x)^2-(4x)^2\cdot\sec^2(4x+y)+1 \
&= (4x)^2-(4x)^2\cdot(1+	an^2(4x+y))+1 \
&= 1-3(4x)^2-y^2 \
2y^2 &= 1-3(4x)^2 \
y^2 &= \frac{1}{2}-\frac{3}{2}(4x)^2 \
\end{align*}

Plugging this into our expression for $\frac{dy}{dx}$, we get:

\begin{align*}
\frac{dy}{dx} &= -\frac{(4x)^2+y^2+8x+1}{(\frac{dy}{dx}+1)(\frac{dy}{dx}+6)} \
&= -\frac{(4x)^2+\frac{1}{2}-\frac{3}{2}(4x)^2+8x+1}{(\frac{dy}{dx}+1)(\frac{dy}{dx}+6)} \
&= \boxed{-\frac{4x^2}{x^2+1}} \
\end{align*}