Question

determine the slope of the line tangent to the curve $xcos(y)=ysin(x)$ at $(\frac{\pi}{4},\frac{\pi}{4})$. $m =$

Explanation:

Step1: Differentiate implicitly

Differentiate $x\cos(y)=y\sin(x)$ with respect to $x$ using product - rule and chain - rule.
The product rule states that $(uv)^\prime = u^\prime v+uv^\prime$.
Differentiating the left - hand side:
Let $u = x$ and $v=\cos(y)$. Then $\frac{d}{dx}(x\cos(y))=\cos(y)-x\sin(y)\frac{dy}{dx}$.
Differentiating the right - hand side:
Let $u = y$ and $v=\sin(x)$. Then $\frac{d}{dx}(y\sin(x))=\frac{dy}{dx}\sin(x)+y\cos(x)$.
So, $\cos(y)-x\sin(y)\frac{dy}{dx}=\frac{dy}{dx}\sin(x)+y\cos(x)$.

Step2: Solve for $\frac{dy}{dx}$

Rearrange the terms to isolate $\frac{dy}{dx}$:
$-x\sin(y)\frac{dy}{dx}-\frac{dy}{dx}\sin(x)=y\cos(x)-\cos(y)$.
Factor out $\frac{dy}{dx}$:
$\frac{dy}{dx}(-x\sin(y)-\sin(x))=y\cos(x)-\cos(y)$.
Then $\frac{dy}{dx}=\frac{\cos(y)-y\cos(x)}{x\sin(y) + \sin(x)}$.

Step3: Substitute the point $x = \frac{\pi}{4},y=\frac{\pi}{4}$

Substitute $x=\frac{\pi}{4}$ and $y = \frac{\pi}{4}$ into $\frac{dy}{dx}$:
$\cos(\frac{\pi}{4})=\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$.
$\frac{dy}{dx}\big|_{x = \frac{\pi}{4},y=\frac{\pi}{4}}=\frac{\frac{\sqrt{2}}{2}-\frac{\pi}{4}\cdot\frac{\sqrt{2}}{2}}{\frac{\pi}{4}\cdot\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}}$.
Factor out $\frac{\sqrt{2}}{2}$ from the numerator and denominator:
$\frac{dy}{dx}\big|_{x = \frac{\pi}{4},y=\frac{\pi}{4}}=\frac{1-\frac{\pi}{4}}{\frac{\pi}{4}+1}=\frac{4 - \pi}{\pi + 4}$.

Answer:

$\frac{4 - \pi}{\pi + 4}$