Question

1. -/1 points

find $\frac{dy}{dx}$ by implicit differentiation.
$ycos(x)=5x^{2}+2y^{2}$
$\frac{dy}{dx}=square$
$xe^{y}=x - y$
$\frac{dy}{dx}=square$

Explanation:

Step1: Differentiate both sides of $xe^y=x - y$

Differentiate left - hand side using product rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x$ and $v = e^y$. The derivative of $x$ with respect to $x$ is $1$, and the derivative of $e^y$ with respect to $x$ is $e^y\frac{dy}{dx}$. The derivative of the right - hand side: the derivative of $x$ is $1$ and the derivative of $y$ is $\frac{dy}{dx}$. So, $e^y+xe^y\frac{dy}{dx}=1-\frac{dy}{dx}$.

Step2: Isolate $\frac{dy}{dx}$ terms

Move all terms with $\frac{dy}{dx}$ to one side: $xe^y\frac{dy}{dx}+\frac{dy}{dx}=1 - e^y$.

Step3: Factor out $\frac{dy}{dx}$

$\frac{dy}{dx}(xe^y + 1)=1 - e^y$.

Step4: Solve for $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{1 - e^y}{1+xe^y}$.

Step5: Differentiate both sides of $y\cos(x)=5x^2 + 2y^2$

Differentiate the left - hand side using product rule: $(y\cos(x))^\prime=y^\prime\cos(x)-y\sin(x)$. Differentiate the right - hand side: the derivative of $5x^2$ is $10x$ and the derivative of $2y^2$ is $4y\frac{dy}{dx}$. So, $\frac{dy}{dx}\cos(x)-y\sin(x)=10x + 4y\frac{dy}{dx}$.

Step6: Isolate $\frac{dy}{dx}$ terms

$\frac{dy}{dx}\cos(x)-4y\frac{dy}{dx}=10x + y\sin(x)$.

Step7: Factor out $\frac{dy}{dx}$

$\frac{dy}{dx}(\cos(x)-4y)=10x + y\sin(x)$.

Step8: Solve for $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{10x + y\sin(x)}{\cos(x)-4y}$.

Answer:

For $xe^y=x - y$, $\frac{dy}{dx}=\frac{1 - e^y}{1+xe^y}$; for $y\cos(x)=5x^2 + 2y^2$, $\frac{dy}{dx}=\frac{10x + y\sin(x)}{\cos(x)-4y}$