Question

work 2 - topics 4 - 5: problem 8 (1 point) if $x e^{9y}+y^{4}sin(8x)=e^{3x}$ defines $y$ implicitly as a differentiable function of $x$ then $\frac{dy}{dx}=square$ note: your answer may contain both $x$ and $y$ variables. preview my answers submit answers you have attempted this problem 0 times. you have 6 attempts remaining.

Explanation:

Step1: Differentiate each term

Differentiate $xe^{9y}$, $y^{4}\sin(8x)$ and $e^{3x}$ with respect to $x$ using product - rule and chain - rule.
The product - rule is $(uv)^\prime = u^\prime v+uv^\prime$ and the chain - rule is $(f(g(x)))^\prime=f^\prime(g(x))g^\prime(x)$.
For $xe^{9y}$:
Let $u = x$ and $v = e^{9y}$. Then $u^\prime=1$ and $v^\prime = 9e^{9y}\frac{dy}{dx}$. So, $(xe^{9y})^\prime=e^{9y}+9xe^{9y}\frac{dy}{dx}$.
For $y^{4}\sin(8x)$:
Let $u = y^{4}$ and $v=\sin(8x)$. Then $u^\prime = 4y^{3}\frac{dy}{dx}$ and $v^\prime=8\cos(8x)$. So, $(y^{4}\sin(8x))^\prime=4y^{3}\sin(8x)\frac{dy}{dx}+8y^{4}\cos(8x)$.
For $e^{3x}$, its derivative is $3e^{3x}$.

Step2: Set up the equation

The derivative of the left - hand side equals the derivative of the right - hand side.
$e^{9y}+9xe^{9y}\frac{dy}{dx}+4y^{3}\sin(8x)\frac{dy}{dx}+8y^{4}\cos(8x)=3e^{3x}$.

Step3: Isolate $\frac{dy}{dx}$

Group the terms with $\frac{dy}{dx}$ on one side:
$9xe^{9y}\frac{dy}{dx}+4y^{3}\sin(8x)\frac{dy}{dx}=3e^{3x}-e^{9y}-8y^{4}\cos(8x)$.
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx}(9xe^{9y}+4y^{3}\sin(8x))=3e^{3x}-e^{9y}-8y^{4}\cos(8x)$.
Then $\frac{dy}{dx}=\frac{3e^{3x}-e^{9y}-8y^{4}\cos(8x)}{9xe^{9y}+4y^{3}\sin(8x)}$.

Answer:

$\frac{3e^{3x}-e^{9y}-8y^{4}\cos(8x)}{9xe^{9y}+4y^{3}\sin(8x)}$