Question

question. find the derivative of y sin(x + 6y)=-7xy - 4 using implicit differentiation. provide your answer below: $\frac{dy}{dx}=$

Explanation:

Step1: Differentiate both sides

Differentiate $y\sin(x + 6y)=-7xy - 4$ with respect to $x$ using product - rule and chain - rule.
The product - rule states that $(uv)^\prime = u^\prime v+uv^\prime$, and the chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$.
For the left - hand side:
Let $u = y$ and $v=\sin(x + 6y)$. Then $\frac{d}{dx}(y\sin(x + 6y))=\frac{dy}{dx}\sin(x + 6y)+y\cos(x + 6y)(1 + 6\frac{dy}{dx})$.
For the right - hand side:
$\frac{d}{dx}(-7xy - 4)=-7y-7x\frac{dy}{dx}$.
So we have $\frac{dy}{dx}\sin(x + 6y)+y\cos(x + 6y)(1 + 6\frac{dy}{dx})=-7y-7x\frac{dy}{dx}$.

Step2: Expand and collect terms with $\frac{dy}{dx}$

Expand the left - hand side: $\frac{dy}{dx}\sin(x + 6y)+y\cos(x + 6y)+6y\cos(x + 6y)\frac{dy}{dx}=-7y-7x\frac{dy}{dx}$.
Collect the terms with $\frac{dy}{dx}$ on one side:
$\frac{dy}{dx}\sin(x + 6y)+6y\cos(x + 6y)\frac{dy}{dx}+7x\frac{dy}{dx}=-7y - y\cos(x + 6y)$.
Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(\sin(x + 6y)+6y\cos(x + 6y)+7x)=-7y - y\cos(x + 6y)$.

Step3: Solve for $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{-7y - y\cos(x + 6y)}{\sin(x + 6y)+6y\cos(x + 6y)+7x}$.

Answer:

$\frac{-7y - y\cos(x + 6y)}{\sin(x + 6y)+6y\cos(x + 6y)+7x}$